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;
int num[500010];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &num[i]);
  multiset<int> s;
  int ct = 0, i = 0, j = 0, mi = 0, mj = 0, ma = 0, dif = 0;
  for (int tt = 0; tt < n; ++tt) {
    if (s.find(num[tt]) == s.end() && dif < k) {
      s.insert(num[tt]);
      ++dif;
      if (tt - i + 1 > ma) {
        mi = i;
        mj = tt;
        ma = tt - i + 1;
      }
    } else if (s.find(num[tt]) == s.end() && dif == k) {
      while (1) {
        int rr = num[i];
        multiset<int>::iterator it = s.find(num[i]);
        s.erase(it);
        ++i;
        if (s.find(rr) == s.end()) break;
      }
      s.insert(num[tt]);
    } else if (s.find(num[tt]) != s.end()) {
      s.insert(num[tt]);
      if (tt - i + 1 > ma) {
        mi = i;
        mj = tt;
        ma = tt - i + 1;
      }
    }
  }
  printf("%d %d", mi + 1, mj + 1);
  return 0;
}
