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 long long int M = 1e6 + 70;
long long int flag[M], flag2[M];
int main() {
  long long int n, k;
  scanf("%lld", &n);
  scanf("%lld", &k);
  long long int a[n + 1];
  long long int i;
  for (i = 0; i < n; i++) scanf("%lld", &a[i]);
  long long int dp[n + 100];
  for (i = n - 1; i > -1; i--) {
    if (flag[a[i]] != 0)
      dp[i] = flag[a[i]];
    else
      dp[i] = 2e9;
    flag[a[i]] = i;
  }
  i = 0;
  long long int j = 0, ansi = 0, ansj = 0, cnt = 0;
  while (i < n && j < n) {
    while (cnt <= k && i < n) {
      if (flag2[a[i]] == 0)
        cnt++, flag2[a[i]] = 1;
      else
        flag2[a[i]]++;
      i++;
    }
    if (cnt > k) i--;
    if (ansi - ansj + 1 < i - j) {
      ansi = i - 1;
      ansj = j;
    }
    if (i < n) {
      while (dp[j] <= i && j < n) j++;
      if (j < n) j++, flag2[a[j - 1]] = 0;
      cnt--;
      i++;
    }
  }
  cout << ansj + 1 << ' ' << ansi + 1 << endl;
  return 0;
}
