Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 2;
const int M = 1e6 + 2;
int a[N];
int cnt[M];
int main() {
  int i, j, n, k, dist = 0;
  int l, r, ans, ansl, ansr;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  for (r = 0; r < n && dist <= k; r++) {
    if (!cnt[a[r]]) ++dist;
    ++cnt[a[r]];
  }
  if (dist > k) {
    --dist;
    --cnt[a[r - 1]];
    --r;
  }
  l = 0;
  ans = r;
  ansl = l;
  ansr = r - 1;
  while (r < n) {
    for (; r < n && dist <= k + 1; r++) {
      if (!cnt[a[r]]) ++dist;
      ++cnt[a[r]];
    }
    if (dist > k + 1) {
      --dist;
      --cnt[a[r - 1]];
      --r;
    }
    for (; l < r && dist > k; l++) {
      --cnt[a[l]];
      if (!cnt[a[l]]) --dist;
    }
    if (r - l > ans) {
      ans = r - l;
      ansl = l;
      ansr = r - 1;
    }
  }
  printf("%d %d\n", ansl + 1, ansr + 1);
  return 0;
}