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;
int a[1000002] = {0};
int las[1000002] = {0};
int nxt[1000002] = {0};
int beg[1000002] = {0};
int n, k, l, r, ansl, ansr = -1, cnt;
int main() {
  int i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!beg[a[i]]) beg[a[i]] = i;
    if (las[a[i]]) nxt[las[a[i]]] = i;
    las[a[i]] = i;
  }
  for (i = 0; i < 1000002; i++) las[i] = 0;
  for (l = 1; l <= n; l++) {
    while (cnt <= k && r <= n) {
      r++;
      if (!las[a[r]]) {
        las[a[r]] = r;
        cnt++;
      }
    }
    las[a[r]] = 0;
    r--;
    cnt--;
    if (r - l > ansr - ansl) {
      ansl = l;
      ansr = r;
    }
    las[a[l]] = nxt[l];
    if (las[a[l]] == 0 || las[a[l]] > r) {
      las[a[l]] = 0;
      cnt--;
    }
  }
  printf("%d %d\n", ansl, ansr);
  return 0;
}
