Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
set<int> s;
int x[maxn];
int n, k;
int main() {
  while (~scanf("%d", &n)) {
    k = 0;
    if (!s.empty()) s.clear();
    for (int i = 1; i <= n; i++) {
      int p;
      scanf("%d", &p);
      if (s.count(p) == 0)
        s.insert(p);
      else {
        x[k++] = i;
        s.clear();
      }
    }
    if (k == 0)
      printf("-1\n");
    else {
      if (x[k - 1] != n) x[k - 1] = n;
      printf("%d\n", k);
      printf("1 %d\n", x[0]);
      for (int i = 0; i < k - 1; i++) printf("%d %d\n", x[i] + 1, x[i + 1]);
    }
  }
  return 0;
}