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[300010], dp[300010], fine[300010], q[300010];
map<int, int> cose;
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &a[i]);
  for (int i = N - 1; i >= 0; i--) {
    if (cose.find(a[i]) == cose.end())
      ;
    else
      fine[i] = cose[a[i]];
    cose[a[i]] = i;
  }
  for (int i = N - 1; i >= 0; i--) {
    dp[i] = ((i + 1 < N) ? dp[i + 1] : 0);
    q[i] = q[i + 1];
    if (fine[i] > i && 1 + dp[fine[i] + 1] > dp[i]) {
      dp[i] = 1 + dp[fine[i] + 1];
      q[i] = fine[i];
    }
  }
  if (dp[0] == 0)
    printf("-1\n");
  else {
    printf("%d\n", dp[0]);
    int i = 0;
    while (i < N) {
      if (dp[q[i] + 1] == 0) {
        printf("%d %d\n", i + 1, N);
        break;
      } else {
        printf("%d %d\n", i + 1, q[i] + 1);
        i = q[i] + 1;
      }
    }
  }
  return EXIT_SUCCESS;
}
