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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 100;
int a[maxn], vis[maxn * 2], num = 0;
void add(int x) {
  vis[a[x]]++;
  if (vis[a[x]] == 1) num++;
}
void del(int x) {
  vis[a[x]]--;
  if (vis[a[x]] == 0) num--;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int ansL = 0, ansR = 0, l = 1, r = 1;
  vis[a[r]] = 1;
  num++;
  while (l <= r && r <= n) {
    if (num <= k) {
      add(++r);
      if (r - l > ansR - ansL) {
        ansL = l;
        ansR = r;
      }
    } else
      del(l++);
  }
  printf("%d %d\n", ansL, ansR - 1);
  return 0;
}
