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[5005554];
int v[1000009];
int k, n, c, mx, ii, jj, p1, p2;
int d() {
  if (c <= k) return false;
  v[a[p1]]--;
  if (v[a[p1]] == 0) {
    c--;
  }
  p1++;
  return c > k;
}
void p() {
  p2++;
  v[a[p2]]++;
  if (v[a[p2]] == 1) {
    c++;
    while (d())
      ;
  }
  if (p2 - p1 > mx) {
    mx = p2 - p1;
    ii = p1;
    jj = p2;
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  ii = 1;
  jj = n;
  mx = -1;
  c = 0;
  p1 = 0;
  p2 = -1;
  while (p2 < n - 1) p();
  cout << ii + 1 << " " << jj + 1;
  return 0;
}
