time limit per test1 second

memory limit per test512 megabytes

inputstandard input

outputstandard output

A boy named Ayrat lives on planet AMI-1511. Each inhabitant of this planet has a talent. Specifically, Ayrat loves running, moreover, just running is not enough for him. He is dreaming of making running a real art.

First, he wants to construct the running track with coating t. On planet AMI-1511 the coating of the track is the sequence of colored blocks, where each block is denoted as the small English letter. Therefore, every coating can be treated as a string.

Unfortunately, blocks aren’t freely sold to non-business customers, but Ayrat found an infinite number of coatings s. Also, he has scissors and glue. Ayrat is going to buy some coatings s, then cut out from each of them exactly one continuous piece (substring) and glue it to the end of his track coating. Moreover, he may choose to flip this block before glueing it. Ayrat want’s to know the minimum number of coating s he needs to buy in order to get the coating t for his running track. Of course, he also want’s to know some way to achieve the answer.

Input

First line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn’t exceed 2100.

Output

The first line should contain the minimum needed number of coatings n or -1 if it’s impossible to create the desired coating.

If the answer is not -1, then the following n lines should contain two integers xi and yi — numbers of ending blocks in the corresponding piece. If xi ≤ yi then this piece is used in the regular order, and if xi > yi piece is used in the reversed order. Print the pieces in the order they should be glued to get the string t.

Examples

input

abc

cbaabc

output

2

3 1

1 3

input

aaabrytaaa

ayrat

output

3

1 1

6 5

8 7

input

ami

no

output

-1

Note

In the first sample string “cbaabc” = “cba” + “abc”.

In the second sample: “ayrat” = “a” + “yr” + “at”.

【题解】



把s1串的所有子串都加入到字典树;在每个节点记录这个节点在s1中的起始位置和终止位置;

加入的时候正串和反串都要加入(加入到同一颗字典树即可);

因为s1串可以多次使用;

所以我们让每次使用都达到最长的距离即可,这样肯定是最优的;

那个from和to域实在设计得巧妙;

从s2串的初始位置开始;从字典树的根节点开始走一条路径即可;

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <iostream>

#include <algorithm>

#include <cstring>

#include <queue>

#include <vector>

#include <stack>

#include <string>

#define lson L,m,rt<<1

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

#define LL long long

using namespace std;

const int MAXN = 4500000;

const int MAXSIZE = 3000;

const int dx[5] = {0,1,-1,0,0};

const int dy[5] = {0,0,0,-1,1};

const double pi = acos(-1.0);

int tree1[MAXN][27],from[MAXN],to[MAXN];

int ans[MAXSIZE][2];

int cnt1 = 0,cnt2 = 0;

string s1,s2;

void add1(int pos,int len)

{

    int now = 0;

    for (int i = pos;i<=len-1;i++)

    {

        int key = s1[i]-'a'+1;

        if (tree1[now][key])

        {

            now = tree1[now][key];

            continue;

        }

        tree1[now][key]=++cnt1;

        now = tree1[now][key];

        from[now] = pos;to[now] = i;

    }

}

void add2(int pos)

{

    int now = 0;

    for (int i = pos;i>=0;i--)

    {

        int key = s1[i]-'a'+1;

        if (tree1[now][key])

        {

            now = tree1[now][key];

            continue;

        }

        tree1[now][key]=++cnt1;

        now = tree1[now][key];

        from[now] = pos;to[now] =i;

    }

}

int main()

{

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

    cin >>s1;

    cin >>s2;

    int len1 = s1.size();

    for (int i = 0;i <= len1-1;i++)

        add1(i,len1);

    for (int i = len1-1;i>=0;i--)

        add2(i);

    int now = 0;

    int len2 = s2.size();

    int dd = 0;

    while (now<=len2-1)

    {

        int key = s2[now]-'a'+1;

        //printf("%d\n",now);

        int temp1 = 0,num1 =now;

        while (num1<=len2-1 && tree1[temp1][key]!=0)

        {

            temp1 = tree1[temp1][key];

            num1++;

            key = s2[num1] - 'a'+1;

        }

        if (num1-1 < now)

        {

            puts("-1");

            return 0;

        }

        else

        {

            dd++;

            now = num1-1;

            ans[dd][0] = from[temp1];

            ans[dd][1] = to[temp1];

            now++;

        }

    }

    printf("%d\n",dd);

    for (int i = 1;i <= dd;i++)

        printf("%d %d\n",ans[i][0]+1,ans[i][1]+1);

    return 0;

}