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 M = 1000000 + 10;
const int INF = int(1e9);
int n, k;
int a[M / 2], d[M];
int best = 1, l = 0, r = 0;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    d[a[i]] = 0;
  }
  d[a[0]]++;
  int diff = 1;
  int l1 = 0, r1 = 0;
  a[n] = a[n - 1];
  while (r1 < n) {
    r1++;
    if (d[a[r1]])
      d[a[r1]]++;
    else {
      diff++;
      d[a[r1]]++;
    }
    if (diff <= k) {
      if (best < r1 - l1 + 1) {
        best = r1 - l1 + 1;
        l = l1;
        r = r1;
        if (r1 == n) best--, r--;
      }
    } else {
      while (diff > k && l1 < r1) {
        if (d[a[l1]] == 1) diff--;
        d[a[l1]]--;
        l1++;
      }
    }
  }
  printf("%d %d\n", l + 1, r + 1);
  return 0;
}
