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;
int main() {
  ios_base::sync_with_stdio(0);
  int n = 0;
  int k = 0;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<pair<int, int>> state;
  map<int, int> d;
  int start = 0;
  int len = 0;
  for (int i = 0; i < n; ++i) {
    d[a[i]]++;
    ++len;
    while (d.size() > k) {
      d[a[start]]--;
      if (d[a[start]] <= 0) d.erase(a[start]);
      ++start;
      --len;
    }
    pair<int, int> curState(start, len);
    state.push_back(curState);
  }
  int resId = 0;
  int maxLen = state[resId].second;
  for (int i = 0; i < n; ++i) {
    if (maxLen < state[i].second) {
      maxLen = state[i].second;
      resId = i;
    }
  }
  const int sRes = state[resId].first + 1;
  const int eRes = sRes + state[resId].second - 1;
  cout << sRes << " " << eRes << endl;
  return 0;
}
