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 comment(linker, "/STACK:64000000")
using namespace std;
const double PI = 6.283185307179586476925286766559;
const int MAXN = 1011000;
int n, k;
int dig[MAXN];
int mas[MAXN];
pair<long long, long long> ans;
int sz = 0;
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> mas[i];
  int dif = 0;
  int l = 0;
  for (int i = 0; i < n; ++i) {
    if (!dig[mas[i]]) {
      dif++;
    }
    dig[mas[i]]++;
    while (dif > k) {
      dig[mas[l]]--;
      if (!dig[mas[l]]) dif--;
      l++;
    }
    if (i - l + 1 > sz) {
      sz = i - l + 1;
      ans = pair<long long, long long>(l, i);
    }
  }
  cout << ans.first + 1 << ' ' << ans.second + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(time(0));
  solve();
  return 0;
}
