time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

A string t is called nice if a string “2017” occurs in t as a subsequence but a string “2016” doesn’t occur in t as a subsequence. For example, strings “203434107” and “9220617” are nice, while strings “20016”, “1234” and “20167” aren’t nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it’s impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits ‘0’–’9’.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

input

8 3

20166766

1 8

1 7

2 8

output

4

3

-1

input

15 5

012016662091670

3 4

1 14

4 15

1 13

10 15

output

-1

2

1

-1

-1

input

4 2

1234

2 4

1 2

output

-1

-1

Note

In the first sample:

In the first query, ugliness(“20166766”) = 4 because all four sixes must be removed.

In the second query, ugliness(“2016676”) = 3 because all three sixes must be removed.

In the third query, ugliness(“0166766”) =  - 1 because it’s impossible to remove some digits to get a nice string.

In the second sample:

In the second query, ugliness(“01201666209167”) = 2. It’s optimal to remove the first digit ‘2’ and the last digit ‘6’, what gives a string “010166620917”, which is nice.

In the third query, ugliness(“016662091670”) = 1. It’s optimal to remove the last digit ‘6’, what gives a nice string “01666209170”.

【题目链接】:http://codeforces.com/contest/750/problem/E

【题解】

    merge矩阵(dp)+线段树;

    m[i][j]表示从状态i->j转移的花费;

    0表示什么都没有

    1表示出现了2

    2表示出现了20

    3表示出现了201

    4表示出现了2017

    出现了2,20，则遇到6的时候我么可以不用管;

    但是如果出现了201或2017,然后又遇到了6,则我们需要把这个6删掉;

    for (i = 0;i <= 4;i++)//除了遇到2,0,1,6,7这5个特别的数字之外,

        m[i][i] = 0;//遇到的时候状态都不会变;所以不用花费(操作);

    if (ch == '2')

    {

        m[0][0] = 1;//什么都没有->什么都没有,删掉这个2

        m[0][1] = 0;//什么都没有->出现了2,花费变成0

        //注意这里m[1][1] = 0,表示从出现了一个2到出现了一个2可以不用删掉任何东西,因为加上一个2也无妨,下面的解释就省略了,希望大家能看得明白.

    }

    if (ch=='0')

    {

        m[1][1] = 1 ;//出现了一个2->出现了一个2,则把0删掉

        m[1][2] = 0;//从出现一个2->出现了20,因为当前就是0,所以什么都不用加

    }

    if (ch=='1')

    {

        ma[2][2] = 1;//出现了20->出现了20,则把1删掉

        ma[2][3] = 0;//出现了20->出现了201，因为刚好遇到一个1则什么都不加

    }

    if (ch=='7')

    {

        m[3][3] = 1;//出现了201->出现了201,则把遇到的7删掉

        m[3][4] = 0;//出现了201->出现了2017,刚好遇到7，则什么都不做

    }

    if (ch=='6')

    {

        a[3][3] = 1;//出现了201->出现了数字6,则一定要把6删掉,不然无法满足题意

        a[4][4] = 1;//出现了2017->出现了数字6,则也一定要把6删掉.

    }

    线段树加一个合并操作就好;

    那个合并的操作和floyd算法类似;

    最后输出m[0][4];表示从什么都没有然后出现"2017"

【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int MAXN = 2e5+10;

const int INF = 7e8;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

struct abc

{

    int m[5][5];

    friend abc operator * (abc a,abc b)

    {

        abc d;

        rep1(i,0,4)

            rep1(j,0,4)

            {

                d.m[i][j] = INF;

                rep1(k,0,4)

                    d.m[i][j] = min(d.m[i][j],a.m[i][k]+b.m[k][j]);

            }

        return d;

    }

};

int n,q;

abc a[MAXN*4];

char s[MAXN];

void build(int l,int r,int rt)

{

    if (l==r)

    {

        char ch = s[l];

        rep1(i,0,4)

            rep1(j,0,4)

                a[rt].m[i][j] = (i==j)?0:INF;

        if (ch == '2')

        {

            a[rt].m[0][0] = 1;

            a[rt].m[0][1] = 0;

        }

        if (ch=='0')

        {

            a[rt].m[1][1] = 1;

            a[rt].m[1][2] = 0;

        }

        if (ch=='1')

        {

            a[rt].m[2][2] = 1;

            a[rt].m[2][3] = 0;

        }

        if (ch=='7')

        {

            a[rt].m[3][3] = 1;

            a[rt].m[3][4] = 0;

        }

        if (ch=='6')

        {

            a[rt].m[3][3] = 1;

            a[rt].m[4][4] = 1;

        }

        return;

    }

    int m = (l+r)>>1;

    build(lson);build(rson);

    a[rt] = a[rt<<1]*a[rt<<1|1];

}

abc query(int L,int R,int l,int r,int rt)

{

    if (L<=l && r <= R)

        return a[rt];

    int m = (l+r)>>1;

    abc temp1,temp2;

    bool f1 = 0,f2 = 0;

    if (L <= m)

    {

        temp1 = query(L,R,lson);

        f1 = 1;

    }

    if (m < R)

    {

        temp2 = query(L,R,rson);

        f2 = 1;

    }

    if (f1&&f2)

        return temp1*temp2;

    if (f1)

        return temp1;

    if (f2)

        return temp2;

}

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    rei(n);rei(q);

    scanf("%s",s+1);

    build(1,n,1);

    rep1(i,1,q)

    {

        int L,R;

        rei(L);rei(R);

        int ans = query(L,R,1,n,1).m[0][4];

        if (ans>=INF)

            puts("-1");

        else

            cout << ans << endl;

    }

    return 0;

}