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 N = 500005;
int a[N], c[2 * N];
int main() {
  int i, n, k;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  int dis = 0, p1 = 0, ans = 0, l = 1, r = 1;
  for (i = 0; i < n; i++) {
    c[a[i]]++;
    if (c[a[i]] == 1) dis++;
    while (dis > k) {
      c[a[p1]]--;
      if (c[a[p1]] == 0) dis--;
      p1++;
    }
    if (dis <= k && i - p1 + 1 > ans) {
      l = p1, r = i;
      ans = i - p1 + 1;
    }
  }
  printf("%d %d", l + 1, r + 1);
}
