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, k;
int a[500005];
map<int, int> m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  set<int> s;
  int i;
  for (i = 1; i <= n; i++) {
    s.insert(a[i]);
    if (s.size() > k) {
      s.erase(a[i]);
      i--;
      break;
    }
    m[a[i]] = i;
  }
  int l = 1, r = i;
  int x = 0;
  for (i = r + 1; i <= n; i++) {
    s.insert(a[i]);
    m[a[i]] = i;
    while (s.size() > k) {
      if (m[a[x]] == x) s.erase(a[x]);
      x++;
    }
    if (i - x + 1 > r - l + 1) {
      r = i;
      l = x;
    }
  }
  cout << l << ' ' << min(n, r) << '\n';
  return 0;
}
