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;
int co[1010000];
int a[1010000];
int n, k, m;
void input() {
  int i;
  m = 0;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (m < a[i]) {
      m = a[i];
    }
  }
  for (i = 0; i <= m; i++) {
    co[i] = 0;
  }
}
void solve() {
  int ans = 0;
  int ax, ay;
  int i, x = 0, z = 0;
  for (i = 0; i < n; i++) {
    if (co[a[i]] == 0) {
      if (z == k) {
        while (x < i) {
          co[a[x]]--;
          if (co[a[x]] == 0) {
            x++;
            break;
          }
          x++;
        }
      } else {
        z++;
      }
    }
    if (i - x + 1 > ans) {
      ax = x;
      ay = i;
      ans = i - x + 1;
    }
    co[a[i]]++;
  }
  printf("%d %d\n", ax + 1, ay + 1);
}
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    input();
    solve();
  }
  return 0;
}
