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 maxN = 1001001;
int a[maxN];
int cnt[maxN];
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  int difs = 0, L = 0, R = 0;
  int ans = -1, lend, rend;
  while (L < n && R < n && L <= R) {
    while (L < n && difs == k) {
      cnt[a[L]]--;
      if (cnt[a[L]] == 0) difs--;
      L++;
    }
    while (R < n && difs <= k) {
      if (cnt[a[R]] == 0 && difs == k) break;
      if (cnt[a[R]] == 0) difs++;
      cnt[a[R]]++;
      R++;
    }
    if (R - L > ans) {
      ans = R - L;
      lend = L;
      rend = R - 1;
    }
  }
  lend++;
  rend++;
  cout << lend << " " << rend << endl;
  return 0;
}
