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 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 n from 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
思路：做了這幾個尺取法的題目，總結(jié)出這樣的一種套路，對于這種類型的尺取來說，這也可以當做一個模板了（和我上一篇博客差別不大）。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=5e5+100; int a[maxx]; int n,k;int main() {scanf("%d%d",&n,&k);for(int i=1;i<=n;i++) scanf("%d",&a[i]);map<int,int> mp;mp.clear();int sum=0,l=1,r=1,_max=0;int L=1,R=1;while(r<=n){while(sum<=k&&r<=n){if(mp[a[r]]==0) sum++;mp[a[r]]++;r++;}if(sum>k) r--,mp[a[r]]--,sum--;if(_max<r-l){_max=r-l;L=l,R=r-1;}while(sum>=k&&l<=n){mp[a[l]]--;if(mp[a[l]]==0) sum--;l++;}}printf("%d %d\n",L,R);return 0; }

努力加油a啊，(^o)/~

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