Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize "trapv"
using namespace std;
int INT_MX = 1e6 + 9;
vector<int> a(26);
int par(int i) {
  while (i != a[i]) i = a[i];
  return i;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  int ans = 0, ans_l = 0;
  int l = 0, r = 0;
  vector<int> m(1000006, 0);
  int count = 0;
  for (int i = 0; i < n; i++) {
    while (r < n) {
      if (m[a[r]] == 0 && count == k) {
        r--;
        break;
      }
      if (m[a[r]] == 0) {
        count++;
        m[a[r]] = 1;
      } else
        m[a[r]]++;
      r++;
    }
    if (r >= n) r = n - 1;
    if (ans < r - i + 1) {
      ans = r - i + 1;
      ans_l = i;
    }
    m[a[i]]--;
    if (m[a[i]] == 0) count--;
    r++;
  }
  cout << ans_l + 1 << " " << ans_l + ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}