time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

There are n pearls in a row. Let’s enumerate them with integers from 1 to n from the left to the right. The pearl number i has the type ai.

Let’s call a sequence of consecutive pearls a segment. Let’s call a segment good if it contains two pearls of the same type.

Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the number of pearls in a row.

The second line contains n integers ai (1 ≤ ai ≤ 109) – the type of the i-th pearl.

Output

On the first line print integer k — the maximal number of segments in a partition of the row.

Each of the next k lines should contain two integers lj, rj (1 ≤ lj ≤ rj ≤ n) — the number of the leftmost and the rightmost pearls in the j-th segment.

Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.

If there are several optimal solutions print any of them. You can print the segments in any order.

If there are no correct partitions of the row print the number “-1”.

Examples

input

5

1 2 3 4 1

output

1

1 5

input

5

1 2 3 4 5

output

-1

input

7

1 2 1 3 1 2 1

output

2

1 3

4 7

【题解】



给你长度为n的序列;

让你把这n个序列分割成最大数目的子段;

使得每个子段里面最少出现两个相同的数字;

用map来判重;

如果之前出现过一次当前扫描到的数字;

则把当前的头尾指针这段区间归为答案;

然后新的头尾指针指向下一个元素

不能直接输出答案，因为最后一段区间的右端点还要指向n;这样才能完成对整个区间的覆盖;

#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 = 4e5;

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 n;

map <int,int>dic;

vector < pair <int,int> > v;

void input_LL(LL &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    LL sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

void input_int(int &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    int sign = 1;

    if (t == '-')sign = -1;

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

    r = r*sign;

}

int main()

{

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

    bool find_ans = 0;

    input_int(n);

    int h = 1,t =1;

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

    {

        int x;

        input_int(x);

        int temp = dic[x];

        if (temp==0)

            dic[x] = 1;

        else

        {

            v.push_back(make_pair(h,t));

            h = i+1,t = i;

            dic.clear();

        }

        t++;

    }

    int len = v.size();

    if (len == 0)

        puts("-1");

    else

    {

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

        for (int i = 0;i <= len-2;i++)

            printf("%d %d\n",v[i].first,v[i].second);

        printf("%d %d\n",v[len-1].first,n);

    }

    return 0;

}