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 main() {
  map<int, pair<int, int> > good;
  int n;
  scanf("%d", &n);
  int a[300004];
  int l[300005];
  int r[300005];
  int k = 0;
  l[k] = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    pair<int, int> element;
    element = good[a[i]];
    if (element.first == 0) {
      good[a[i]].first = 1;
      good[a[i]].second = i;
      continue;
    }
    if (element.first == 1) {
      if (k == 0 || r[k - 1] < element.second) {
        l[k + 1] = i + 1;
        r[k] = i;
        k++;
        good[a[i]].first = 0;
        good[a[i]].second = 0;
      } else if (k > 0 && r[k - 1] > element.second) {
        good[a[i]].first = 1;
        good[a[i]].second = i;
      }
    }
  }
  if (k == 0) {
    printf("-1\n");
    return 0;
  }
  if (r[k - 1] < n - 1) r[k - 1] = n - 1;
  printf("%d\n", k);
  for (int i = 0; i < k; i++) {
    printf("%d %d\n", l[i] + 1, r[i] + 1);
  }
  return 0;
}
