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 n;
vector<int> vans;
set<int> s;
int main() {
  scanf("%d", &n);
  int start = 1;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    if (s.count(x)) {
      vans.push_back(start);
      vans.push_back(i);
      start = i + 1;
      s.clear();
    } else {
      s.insert(x);
    }
  }
  if (vans.empty()) {
    printf("-1\n");
    return 0;
  } else {
    printf("%d\n", vans.size() / 2);
    for (int i = 0; i < vans.size() - 2; i += 2)
      printf("%d %d\n", vans[i], vans[i + 1]);
    if (!s.empty())
      printf("%d %d\n", vans[vans.size() - 2], n);
    else
      printf("%d %d\n", vans[vans.size() - 2], vans[vans.size() - 1]);
  }
  return 0;
}
