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;
const int mx = 5 * 1e5 + 100;
const int mn = 1e6 + 100;
int v[mx];
int u[mn];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0, _ = n; i < _; i++) {
    scanf("%d", &v[i]);
  }
  int x = 0, l = 0, d = 0, ansl = 0, ansr = 0;
  for (int i = 0, _ = n; i < _; i++) {
    if (!u[v[i]]) x++;
    u[v[i]]++;
    while (x > k) {
      u[v[l]]--;
      if (!u[v[l]]) x--;
      l++;
    }
    if (d < i - l + 1) {
      d = i - l + 1;
      ansl = l;
      ansr = i;
    }
  }
  printf("%d %d\n", ansl + 1, ansr + 1);
}