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 long long OO = 1e9;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  map<int, int> mp;
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    if (mp[a[i]]) {
      ans.push_back(i + 1);
      mp.clear();
    } else
      mp[a[i]] = i + 1;
  }
  if (ans.size() == 0) {
    cout << -1 << endl;
    return 0;
  }
  printf("%d\n", (int)ans.size());
  int h = 1;
  for (int i = 0; i < (int)ans.size(); i++) {
    if (i + 1 == ans.size()) ans[i] = n;
    printf("%d %d\n", h, ans[i]);
    h = ans[i] + 1;
  }
}
