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 a[300010];
vector<pair<int, int> > r(300010);
map<int, int> mp;
int main() {
  cin.tie(NULL);
  int n;
  scanf("%d", &n);
  int i, sz = 0, start;
  for (i = 0; i < n; i++) {
    scanf("%d", (a + i));
  }
  mp[a[0]]++;
  start = 0;
  for (i = 1; i < n; i++) {
    mp[a[i]]++;
    if (mp[a[i]] == 2) {
      r[sz++] = (make_pair(start + 1, i + 1));
      start = i + 1;
      mp.clear();
    }
  }
  if (sz == 0)
    cout << -1 << endl;
  else {
    r[sz - 1].second = n;
    cout << sz << endl;
    for (i = 0; i < sz; i++) {
      printf("%d %d\n", r[i].first, r[i].second);
    }
  }
  return 0;
}
