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 main() {
  int n, k;
  scanf("%d %d\n", &n, &k);
  vector<int> a(n);
  for (auto&& ax : a) scanf("%d", &ax);
  int ans_l = 0, ans_r = 0;
  map<int, int> ma;
  int r = 0;
  for (int l = 0; l < n; ++l) {
    while (r < n && (ma.size() < k || ma.count(a[r]))) {
      ++ma[a[r]];
      ++r;
    }
    if (r - l > ans_r - ans_l) ans_r = r, ans_l = l;
    --ma[a[l]];
    if (ma[a[l]] == 0) ma.erase(a[l]);
  }
  printf("%d %d\n", ans_l + 1, ans_r);
  return 0;
}
