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 = 2e6;
int a[maxn], b[maxn], n, j, cur, res1, res2, k;
void add(int u) {
  b[a[u]]++;
  if (b[a[u]] == 1) cur++;
}
void rem(int u) {
  b[a[u]]--;
  if (!b[a[u]]) cur--;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  j = 1;
  for (int i = 1; i <= n; i++) {
    while (j <= n and cur <= k) add(j++);
    if (cur > k) rem(--j);
    if ((j - 1 - i) >= (res2 - res1)) res1 = i, res2 = j - 1;
    rem(i);
  }
  cout << res1 << " " << res2;
}
