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 = 500001, M = 1000001;
int a[N], b[M];
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  memset(b, 0, sizeof b);
  int start = 0, end = 0, startb = start, endb = end, cnt = 1;
  b[a[0]] = 1;
  for (int i = 1; i < n; i++) {
    if ((cnt == k && b[a[i]] != 0) || cnt < k) {
      if (b[a[i]] == 0) cnt++;
      b[a[i]]++;
      end++;
    } else {
      if (end - start > endb - startb) {
        endb = end;
        startb = start;
      }
      if (b[a[start]] == 1) cnt--;
      b[a[start]]--;
      start += 1;
      i--;
    }
  }
  if (end - start > endb - startb) {
    endb = end;
    startb = start;
  }
  cout << startb + 1 << " " << endb + 1 << "\n";
}
