CodeForces

The array a with n integers is given. Let's call the sequence of one or more consecutive elements ina segment. Also let's call the segmentk-good if it contains no more thank 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 usescanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·10⁵) — the number of elements ina and the parameter k.

The second line contains n integers a_i (0 ≤ a_i ≤ 10⁶) — the elements of the arraya.

Output

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

Example

Input

5 51 2 3 4 5

Output

1 5

Input

9 36 5 1 2 3 2 1 4 5

Output

3 7

Input

3 11 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;}