Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[1000001], p1[1000001] = {0}, p2[1000001] = {0}, prev_index[1000001],
                next_index[1000001], dp[1000001];
int main() {
  int n, k, i, j, temp;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    dp[i] = 1;
  }
  for (i = 1; i <= n; i++) {
    if (p1[a[i]] == 0)
      prev_index[i] = i;
    else if (p1[a[i]] > 0)
      prev_index[i] = p1[a[i]];
    p1[a[i]] = i;
  }
  for (j = n; j >= 1; j--) {
    if (p2[a[j]] == 0)
      next_index[j] = j;
    else if (p2[a[j]] > 0)
      next_index[j] = p2[a[j]];
    p2[a[j]] = j;
  }
  i = 1, j = 1;
  int cnt = 1, flag = 1, c = 0, le = 1, ri = 1, ma = -1;
  while (j != n) {
    if ((j + 1) <= n) {
      if (flag == 1) {
        if (prev_index[j + 1] >= i && prev_index[j + 1] <= j)
          dp[i]++;
        else if (prev_index[j + 1] == j + 1 || prev_index[j + 1] < i)
          cnt++, dp[i]++;
        if (cnt > k)
          flag = 0, c = 0, cnt--, dp[i]--;
        else
          j++;
        if (j == n) flag = 0;
      }
    }
    if (cnt <= k && flag == 0) {
      if (dp[i] > ma) ma = dp[i], le = i, ri = j;
      if (c == 0) i++, c = 1;
      if (next_index[i - 1] >= i && next_index[i - 1] <= j)
        dp[i] = dp[i - 1] - 1, i++;
      else if (next_index[i - 1] == i - 1 || next_index[i - 1] > j)
        flag = 1, cnt--, dp[i] = dp[i - 1] - 1;
    }
    if (cnt == k - 1 && j == n) break;
  }
  cout << le << " " << ri;
  return 0;
}