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 int MOD = 1e9 + 7;
const long double PI = acos((long double)-1);
const int maxN = 5e5 + 1;
int arr[maxN], n, k, l, r;
bool check(int len) {
  vector<int> freq((int)1e6 + 1);
  int cnt = 0;
  for (int i = 1; i <= len; ++i) {
    freq[arr[i]]++;
    if (freq[arr[i]] == 1) ++cnt;
  }
  if (cnt <= k) {
    l = 1, r = len;
    return true;
  }
  for (int i = len + 1; i <= n; ++i) {
    freq[arr[i - len]]--;
    if (freq[arr[i - len]] == 0) {
      --cnt;
    }
    freq[arr[i]]++;
    if (freq[arr[i]] == 1) ++cnt;
    if (cnt <= k) {
      r = i, l = r - len + 1;
      return true;
    }
  }
  return false;
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
  }
  int s = 0, e = n + 1;
  while (s + 1 < e) {
    int m = (s + e) / 2;
    if (check(m)) {
      s = m;
    } else {
      e = m;
    }
  }
  cout << l << " " << r << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
