CodeForces - 616D Longest k-Good Segment （尺取法）

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-good if 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 a_i (0 ≤ a_i ≤ 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 n from left to right.

Example

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

题目大意就是给一个数组，找到最长的区间，使得不同元素的个数为k

今天才知道尺取法是什么意思，“尺”就是一个不定长的区间，然后一直取，取到取不下区间起点就向左移，还能取就区间终点向后移

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int num[1000005];
int vis[1000005];
int main()
{
    int n,k;
    int sta;
    while(~scanf("%d%d",&n,&k))
    {
        memset(vis,0,sizeof vis);
        for(int i=1;i<=n;i++)  scanf("%d",&num[i]);
        int cnt=0;
        int maxlen=0;
        int tempsta=1;
        int i=1;
        while(i<=n&&tempsta<=n-k+1)
        {
            while(cnt<=k&&i<=n)
            {
                if(!vis[num[i]])
                   cnt++;
                vis[num[i]]++;
                i++;
            }
            i--;
            vis[num[i]]--;
            cnt--;
            if(cnt<k)
                i++;
            if(maxlen<i-tempsta)
            {
                sta=tempsta;
                maxlen=i-tempsta;
            }
            if(cnt<k)
                i--;
            vis[num[tempsta]]--;
            if(vis[num[tempsta]]==0)
                cnt--;
            tempsta++;
        }
        printf("%d %d\n",sta,sta+maxlen-1);
    }
     return 0;
}