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 = 1e6 + 6;
int ara[N];
int fre[N];
int l, r, i, j, cnt, len;
int main() {
  int n, m;
  cin >> n >> m;
  for (i = 0; i < n; i++) cin >> ara[i];
  for (int i = 0; i < n; i++) {
    fre[ara[i]]++;
    if (fre[ara[i]] == 1) cnt++;
    while (cnt > m) {
      fre[ara[j]]--;
      if (fre[ara[j]] == 0) cnt--;
      j++;
    }
    if (i - j > len) {
      len = i - j;
      r = i;
      l = j;
    }
  }
  cout << l + 1 << " " << r + 1 << endl;
}
