Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 3e5;
const int INF_ = 0x3f3f3f3f;
int a[MAX_N + 10];
int main() {
  int n;
  scanf("%d", &n);
  int b = 1;
  vector<pair<int, int>> ans;
  set<int> accur;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (accur.count(a[i]) > 0) {
      accur.clear();
      ans.push_back(make_pair(b, i));
      b = i + 1;
    } else {
      accur.insert(a[i]);
    }
  }
  if (ans.size() == 0) {
    printf("-1\n");
  } else {
    printf("%d\n", ans.size());
    ans.back().second = n;
    for (int i = 0; i < ans.size(); i++) {
      printf("%d %d\n", ans[i].first, ans[i].second);
    }
  }
}