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 + 2;
int a[N], b[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int i, j, n, k, mx = 0, l = 1, x, y, c = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (b[a[i]] == 0) {
      c++;
    }
    b[a[i]]++;
    while (c > k) {
      b[a[l]]--;
      if (b[a[l]] == 0) {
        c--;
      }
      l++;
    }
    if (c <= k) {
      if ((i - l + 1) > mx) {
        mx = i - l + 1;
        x = l, y = i;
      }
    }
  }
  cout << x << " " << y << "\n";
  return 0;
}
