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[1000000];
int b[1000000];
int main() {
#pragma warning(disable : 4996)
  int n, k, dl, kol;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]] = 0;
  }
  int r = 1;
  int otl = 1;
  int otr = 1;
  int otdl = 1;
  bool f = false;
  kol = 1;
  dl = 1;
  b[a[1]] = 1;
  for (int l = 1; l <= n; l++) {
    f = false;
    while (kol <= k && r < n && f == false) {
      if (kol == k) {
        if (b[a[r + 1]] != 0) {
          dl++;
          r++;
          b[a[r]]++;
        } else {
          f = true;
        }
      } else {
        r++;
        dl++;
        b[a[r]]++;
        if (b[a[r]] == 1) kol++;
      }
    }
    if (dl > otdl) {
      otl = l;
      otr = r;
      otdl = dl;
    }
    if (b[a[l]] == 1) kol--;
    dl--;
    b[a[l]]--;
  }
  cout << otl << " " << otr << " ";
  return 0;
}
