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;
const int inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
int n, k, a[500005], t[1000005], r, cnt, ans, s1, s2;
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++) scanf("%d", a + i);
  r = 0;
  for (int(i) = (1); (i) <= (n); (i)++) {
    if (i > 1) {
      t[a[i - 1]]--;
      if (t[a[i - 1]] == 0) cnt--;
    }
    while (cnt <= k && r < n) {
      r++;
      if (t[a[r]] == 0) cnt++;
      t[a[r]]++;
    }
    if (cnt > k) {
      t[a[r]]--;
      if (t[a[r]] == 0) cnt--;
      r--;
    }
    if (ans < r - i + 1) ans = r - i + 1, s1 = i, s2 = r;
  }
  printf("%d %d\n", s1, s2);
  return 0;
}
