Longest k-Good Segment CodeForces - 616D

The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-goodif it contains no more than k different values.

Find any longest k-good segment.

As the 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 two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to nfrom left to right.

Examples

Input

5 5
1 2 3 4 5

Output

1 5

Input

9 3
6 5 1 2 3 2 1 4 5

Output

3 7

Input

3 1
1 2 3

Output

1 1

AC代码（尺取法维护一个符合要求的区间）

Select Code

#include <iostream>
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
 
int s[500030],vis[1000030];
 
int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (int i=0;i<n;i++)
            scanf("%d",&s[i]);
 
        memset(vis,0,sizeof(vis));
        int pos=0,maxx=0,ansl=0,ansr=0,num=0;
        for (int i=0;i<n;i++)
        {
            if (vis[s[i]]==0)
                num++;
            vis[s[i]]++;
 
            while (num>m)
            {
                vis[s[pos]]--;
                if (vis[s[pos]]==0) num--;
                pos++;
            }
            if (maxx<i-pos+1)
            {
                maxx=i-pos+1;
                ansl=pos;
                ansr=i;
            }
        }
        printf("%d %d\n",ansl+1,ansr+1);
    }
}