Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[300000 + 100], l[300000 + 100], r[300000 + 100];
int read() {
  int _ = 0, ___ = 1;
  char __ = getchar();
  while (__ < '0' || __ > '9') {
    if (__ == '-') ___ = -1;
    __ = getchar();
  }
  while (__ >= '0' && __ <= '9') {
    _ = _ * 10 + __ - '0';
    __ = getchar();
  }
  return _ * ___;
}
int main() {
  int n, pos, num;
  map<int, int> Map;
  while (cin >> n) {
    num = 0;
    pos = 1;
    Map.clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (Map[a[i]] != 0) {
        l[num] = pos;
        r[num++] = i;
        pos = i + 1;
        Map.clear();
      } else {
        Map[a[i]] = 1;
      }
    }
    r[num - 1] = n;
    if (num) {
      printf("%d\n", num);
    } else {
      printf("-1\n");
    }
    for (int i = 0; i < num; i++) {
      printf("%d %d\n", l[i], r[i]);
    }
  }
  return 0;
}