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;
const int maxn = 3e5 + 2;
long long n, a[maxn];
map<long long, long long> vis;
long long l[maxn], r[maxn];
int main() {
  scanf("%lld", &n);
  long long ans = 0, t = 1;
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    if (!vis[a[i]]) {
      vis[a[i]] = 1;
    } else {
      vis[a[i]]++;
      l[ans] = t;
      r[ans++] = i;
      t = i + 1;
      vis.clear();
    }
  }
  r[ans - 1] = n;
  if (!ans) {
    printf("-1\n");
    return 0;
  }
  printf("%lld\n", ans);
  for (long long i = 0; i < ans; i++) {
    printf("%lld %lld\n", l[i], r[i]);
  }
}
