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() {
  ios_base::sync_with_stdio(0);
  int beg = 0, end = 0, l = 0, r = 0, a, b, ans = 1;
  scanf("%d %d", &a, &b);
  set<int> kn;
  int v[int(1e6 + 1)], onthe[int(1e6 + 1)];
  memset(v, 0, sizeof v);
  for (int i = 0; i < a; ++i) {
    scanf("%d", &onthe[i]);
    kn.insert(onthe[i]);
    ++v[onthe[i]];
    if (kn.size() <= b) {
      if (end - beg + 1 > ans) {
        l = beg;
        r = end;
        ans = end - beg + 1;
      }
    } else {
      while (kn.size() > b) {
        if (v[onthe[beg]] <= 1) kn.erase(kn.find(onthe[beg]));
        --v[onthe[beg]];
        ++beg;
      }
    }
    ++end;
  }
  printf("%d %d\n", l + 1, r + 1);
  return 0;
}
