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;
signed main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int b[1000001], c[1000001];
  for (int i = 0; i < 1000001; i++) b[i] = 0, c[i] = -1;
  int l = 0;
  int r = 0;
  int m = 0;
  int cur = 0;
  pair<int, int> d[n];
  set<pair<int, int> > s;
  for (int i = 0; i < n; i++) {
    if (b[a[i]] == 0) {
      b[a[i]] = 1;
      c[a[i]] = i;
      if (cur < k) {
        cur += 1;
        r = i;
      } else {
        pair<int, int> p = *(s.begin());
        b[p.second] = 0;
        s.erase(p);
        l = p.first + 1;
        r = i;
      }
      s.insert({i, a[i]});
    } else {
      r = i;
      b[a[i]] += 1;
      s.erase({c[a[i]], a[i]});
      s.insert({i, a[i]});
      c[a[i]] = i;
    }
    d[i] = {l, r};
  }
  for (int i = 0; i < n; i++) {
    if ((d[i].second - d[i].first + 1) > (d[m].second - d[m].first + 1)) {
      m = i;
    }
  }
  cout << d[m].first + 1 << " " << d[m].second + 1 << "\n";
  return 0;
}
