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 double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int maxn = 5e5 + 10;
int n, k;
int a[maxn];
int vis[(int)1e6 + 10];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  int maxl, maxr;
  int maxlen = 0;
  maxl = 1;
  maxr = 1;
  int l, r;
  l = 1;
  vis[a[1]] = 1;
  int len;
  len = 1;
  int cnt;
  cnt = 1;
  for (int r = 2; r <= n; r++) {
    if (!vis[a[r]]) {
      cnt++;
    }
    vis[a[r]]++;
    if (cnt <= k && r - l + 1 > maxlen) {
      maxlen = r - l + 1;
      maxl = l;
      maxr = r;
    } else
      while (cnt > k) {
        vis[a[l]]--;
        if (!vis[a[l]]) {
          cnt--;
        }
        l++;
      }
  }
  printf("%d %d\n", maxl, maxr);
  return 0;
}