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;
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
struct greaterr {
  template <class T>
  bool operator()(T const &a, T const &b) const {
    return a > b;
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<int, int> mp;
  int i;
  for (i = 0; i < n; i++) {
    if (int((mp).size()) > k) {
      break;
    }
    if (mp.find(a[i]) == mp.end()) {
      if (int((mp).size()) < k) {
        mp[a[i]] = 1;
      } else {
        break;
      }
    } else {
      mp[a[i]]++;
    }
  }
  int l = 0, r = i - 1;
  int diff = r - l, maxdiff = r - l, maxl = l, maxr = r;
  i = r;
  int z = 0;
  std::map<int, int>::iterator itr;
  while (r < n) {
    int ab = 0;
    if (l == r && r == n - 1) break;
    if (mp[a[l]] >= 2) {
      mp[a[l]]--;
      l = l + 1;
    } else {
      mp.erase(mp.find(a[l]));
      l = l + 1;
    }
    for (int j = r + 1; j < n; j++) {
      if (mp.size() > k) break;
      if (mp.find(a[j]) != mp.end()) {
        r = j;
        mp[a[j]]++;
      } else {
        if (int((mp).size()) < k) {
          r = j;
          mp[a[j]]++;
        } else {
          break;
        }
      }
    }
    diff = r - l;
    if (diff > maxdiff) {
      maxl = l;
      maxr = r;
      maxdiff = diff;
    }
  }
  cout << maxl + 1 << " " << maxr + 1 << "\n";
  return 0;
}
