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 = 1111122;
const int M = 200005;
int a[N], v[N];
int main() {
  int m, i, j, n, k, r, l, maxx = 0, s = 0, r1, l1;
  cin >> n >> k;
  memset(v, 0, sizeof(v));
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  r = 1;
  l = 1;
  while (1) {
    if (r > n) break;
    while (s < k) {
      if (!v[a[r]]) {
        s++;
        v[a[r]]++;
        r++;
      } else {
        v[a[r]]++;
        r++;
      }
      if (r > n) break;
    }
    if (maxx < r - l) {
      maxx = r - l;
      l1 = l;
      r1 = r - 1;
    }
    if (r > n) break;
    if (s >= k) {
      if (v[a[r]] > 0) {
        v[a[r]]++;
        r++;
        if (maxx < r - l) {
          maxx = r - l;
          l1 = l;
          r1 = r - 1;
        }
      } else {
        if (v[a[l]] > 1) {
          v[a[l]]--;
          l++;
        } else if (v[a[l]] == 1) {
          s--;
          v[a[l]]--;
          l++;
        }
      }
    }
  }
  cout << l1 << " " << r1 << endl;
  return 0;
}
