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 inf = (int)1e9;
const double pi = acos(-1.0);
const double eps = 1e-9;
int a[2000100], n, w[2000100], cnt, l, k, res;
void add(int x) {
  w[x]++;
  if (w[x] == 1) cnt++;
}
void del(int x) {
  w[x]--;
  if (w[x] == 0) cnt--;
}
int resi = 0;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  l = 1;
  for (int i = 1; i <= n; i++) {
    add(a[i]);
    while (cnt > k) {
      del(a[l]);
      l++;
    }
    res = max(res, i - l + 1);
    if (res == i - l + 1) {
      resi = l;
    }
  }
  cout << resi << " " << resi + res - 1;
  return 0;
}
