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;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long N = 2e5 + 100, K = 31;
const double pi = 3.14159265358979323846264338327950288419716939937510;
void solve(long long ttt) {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  unordered_map<long long, long long> mp;
  long long ansl = 0, ansr = 0, ans = 0;
  for (long long l = 0, r = 0; r < n; ++r) {
    ++mp[a[r]];
    while (mp.size() > k) {
      --mp[a[l]];
      if (mp[a[l]] == 0) mp.erase(a[l]);
      ++l;
    }
    if (r - l + 1 > ans) {
      ans = r - l + 1;
      ansl = l + 1;
      ansr = r + 1;
    }
  }
  cout << ansl << " " << ansr;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long ttt = 1;
  while (ttt--) {
    solve(ttt);
    cout << '\n';
  }
}
