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 const maxn = 5e5 + 100;
int const maxN = 1e6 + 100;
int a[maxn];
int f[maxN];
int main() {
  int n, k;
  scanf(" %d %d", &n, &k);
  int tam = -0x3f3f3f3f, l = 1, r = n;
  for (int i = 0; i < n; ++i) scanf(" %d", a + i);
  int T = 0;
  for (int j = 0, i = 0; j < n; ++j) {
    while (n > i && k >= T + (f[a[i]] == 0)) {
      f[a[i]]++;
      T += (f[a[i]] == 1);
      i++;
    }
    if (T == k && tam < (i - j)) {
      tam = i - j;
      l = j + 1;
      r = i;
    }
    T -= (--f[a[j]] == 0);
    if (n <= i) break;
  }
  return !printf("%d %d\n", l, r);
}
