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>
int num[500009];
int vis[1000009];
int n, m;
int maxx, left, right, tmp;
void solve() {
  int differ = 0;
  int cur = 0;
  for (int i = 1; i <= n; i++) {
    int val = differ;
    if (val < m) {
      while (1) {
        if (++cur > n || val == m && !vis[num[cur]]) break;
        if (vis[num[cur]]) {
          vis[num[cur]]++;
          continue;
        } else if (!vis[num[cur]]) {
          val++;
          vis[num[cur]] = 1;
          continue;
        }
      }
      cur = cur - 1;
      differ = val;
    }
    tmp = cur - i + 1;
    if (vis[num[i]] == 1) {
      vis[num[i]] = 0;
      differ--;
    } else if (vis[num[i]] > 1)
      vis[num[i]]--;
    if (tmp > maxx) {
      maxx = tmp;
      left = i;
      right = cur;
    }
  }
}
int main(void) {
  scanf("%d%d", &n, &m);
  memset(vis, 0, sizeof(vis));
  left = 0;
  right = 0;
  maxx = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  solve();
  printf("%d %d\n", left, right);
  return 0;
}
