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;
const int MAX = 1000000;
struct s {
  int l, r;
} a[MAX];
int n;
map<long long, int> mp;
int main() {
  scanf("%d", &n);
  mp.clear();
  long long t;
  int sum = 0;
  int left = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t);
    mp[t]++;
    if (mp[t] == 2) {
      a[sum].l = left;
      a[sum].r = i;
      sum++;
      left = i + 1;
      mp.clear();
    } else
      mp[t] = 1;
  }
  if (a[sum - 1].r < n) a[sum - 1].r = n;
  if (sum > 0) {
    printf("%d\n", sum);
    for (int i = 0; i < sum; i++) {
      printf("%d %d\n", a[i].l, a[i].r);
    }
  } else
    printf("-1\n");
  return 0;
}