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;
long long a[500005], ansl, ansr, len, k, n, cnt[1000005], use;
bool check(long long dl) {
  use = 0;
  for (long long i = 1; i <= n; i++) {
    if (cnt[a[i]] == 0) use++;
    cnt[a[i]]++;
    if (i > dl) {
      cnt[a[i - dl]]--;
      if (cnt[a[i - dl]] == 0) use--;
    }
    if (i >= dl) {
      if (use <= k) {
        if (dl > len) {
          ansl = i - dl + 1;
          ansr = i;
          len = dl;
          for (long long j = i; j >= max(1LL, i - dl); j--) cnt[a[j]] = 0;
          return true;
        }
      }
    }
  }
  for (long long i = n; i >= max(1LL, n - dl); i--) cnt[a[i]] = 0;
  return false;
}
int main() {
  long long i, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  long long l = 1, r = n;
  check(n);
  while (r - l > 0) {
    long long mid = (l + r) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  for (i = max(r - 2, 0LL); i <= min(n, r + 2); i++) check(i);
  cout << ansl << " " << ansr;
}
