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;
struct mapi {
  vector<long long> a;
  long long ct;
  mapi(long long n) {
    a = vector<long long>(n + 1, 0);
    ct = 0;
  }
  void add(long long x) {
    if (++a[x] == 1) ct++;
  }
  void del(long long x) {
    if (--a[x] == 0) ct--;
  }
};
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  long long l = 0;
  mapi mp(1000000);
  long long r = 0;
  long long a, b;
  long long ans = -1;
  while (r < n) {
    if (mp.ct <= k) {
      if ((r - l) > ans) {
        ans = r - l;
        a = l;
        b = r - 1;
      }
    }
    if (mp.ct <= k || r == l) {
      mp.add(v[r]);
      r++;
    } else {
      mp.del(v[l]);
      l++;
    }
  }
  if (mp.ct <= k) {
    if ((r - l + 1) > ans) {
      ans = r - l + 1;
      a = l;
      b = r - 1;
    }
  }
  cout << a + 1 << " " << b + 1;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}
