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;
int n, a[1000009] = {};
queue<int> q;
int main() {
  int l = 0, k, mx = 1, mxl = 0, x, c = 0, u = 0;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    q.push(x);
    if (!a[x]) c++;
    a[x]++;
    while (c > k) {
      l++;
      u = q.front();
      q.pop();
      a[u]--;
      if (a[u] == 0) c--;
    }
    if (i - l + 1 > mx) mxl = l, mx = i - l + 1;
  }
  printf("%d %d\n", mxl + 1, mx + mxl);
  return 0;
}
