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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long N = 1000005;
long long cnt[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ai = 0, aj = 0;
  long long cur = 0, le = 0, ri = 0, mx = 0;
  while (ri < n) {
    if (!cnt[a[ri]]) {
      cur++;
      cnt[a[ri]]++;
      if (cur <= k) {
        if (ri - le + 1 >= mx) {
          ai = le;
          aj = ri;
          mx = ri - le + 1;
        }
        ri++;
      } else {
        cnt[a[le]]--;
        if (!cnt[a[le]]) cur--;
        le++;
        cnt[a[ri]]--;
        if (!cnt[a[ri]]) cur--;
      }
    } else {
      if (ri - le + 1 >= mx) {
        ai = le;
        aj = ri;
        mx = ri - le + 1;
      }
      cnt[a[ri]]++;
      ri++;
    }
  }
  cout << ai + 1 << " " << aj + 1 << "\n";
  return 0;
}
