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;
void solve();
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}
int const MAX = 1e6 + 7;
int cnt[MAX];
void solve() {
  int n, k;
  cin >> n >> k;
  int val[n];
  for (int i = 0; i < n; i++) cin >> val[i];
  int x, y;
  int ans = 0;
  int l = 0;
  int diff = 0;
  for (int r = 0; r < n; r++) {
    if (++cnt[val[r]] == 1) {
      diff++;
    }
    while (diff > k) {
      if (--cnt[val[l]] == 0) diff--;
      l++;
    }
    if (r - l + 1 > ans) {
      ans = r - l + 1;
      x = l;
      y = r;
    }
  }
  cout << x + 1 << " " << y + 1 << endl;
}
