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 mn = 5e5 + 5;
const int mod = 1e9 + 7;
int n, k;
int a[mn];
int ansl, ansr, anslen = -1;
int cnt[mn * 2];
int typecnt;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l, r = 0;
  for (int l = 1; l <= n; l++) {
    while (r < n) {
      if (typecnt == k && cnt[a[r + 1]] == 0) break;
      r++;
      cnt[a[r]]++;
      if (cnt[a[r]] == 1) typecnt++;
    }
    if (r - l > anslen) {
      ansl = l, ansr = r;
      anslen = r - l;
    }
    cnt[a[l]]--;
    if (cnt[a[l]] == 0) typecnt--;
  }
  printf("%d %d\n", ansl, ansr);
}
