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>
using namespace std;
int a[500005], cnt[1000005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  int n, k, A = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int l = 0, r = n, ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, A = 0;
    for (int i = 1; i <= mid; i++) {
      cnt[a[i]]++;
      if (cnt[a[i]] == 1) A++;
    }
    bool q = 0;
    if (A <= k) q = 1;
    for (int i = 2; i + mid - 1 <= n; i++) {
      if (q) break;
      cnt[a[i - 1]]--;
      if (cnt[a[i - 1]] == 0) A--;
      cnt[a[i + mid - 1]]++;
      if (cnt[a[i + mid - 1]] == 1) A++;
      if (A <= k) {
        q = 1;
        break;
      }
    }
    if (q)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
    for (int i = 1; i <= n; i++) cnt[a[i]] = 0;
  }
  A = 0;
  for (int i = 1; i <= n; i++) cnt[a[i]] = 0;
  for (int i = 1; i <= ans; i++) {
    cnt[a[i]]++;
    if (cnt[a[i]] == 1) A++;
  }
  if (A <= k) {
    cout << 1 << " " << ans;
    return 0;
  }
  for (int i = 2; i + ans - 1 <= n; i++) {
    cnt[a[i - 1]]--;
    if (cnt[a[i - 1]] == 0) A--;
    cnt[a[i + ans - 1]]++;
    if (cnt[a[i + ans - 1]] == 1) A++;
    if (A <= k) {
      cout << i << " " << i + ans - 1;
      return 0;
    }
  }
}