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() {
  int n, a;
  scanf("%d", &n);
  map<int, pair<int, int> > ansmap;
  map<int, int> cmap;
  int k = 0;
  ansmap[++k] = make_pair(1, n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    if (cmap[a] == 1) {
      ansmap[k] = make_pair(ansmap[k].first, i);
      cmap.clear();
      ansmap[++k] = make_pair(i + 1, n);
    } else
      cmap[a] = 1;
  }
  k = ansmap.size() - 1;
  if (k == 0) {
    printf("-1\n", k);
  } else {
    printf("%d\n", k);
    if (ansmap[k].second != n) ansmap[k].second = n;
    for (int i = 1; i <= k; i++)
      printf("%d %d\n", ansmap[i].first, ansmap[i].second);
  }
  return 0;
}
