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;
int inp[300010];
set<int> second;
vector<int> V;
int main() {
  int n, i, res = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", inp + i);
  }
  V.push_back(0);
  for (i = 0; i < n; i++) {
    if (second.find(inp[i]) != second.end()) {
      res++;
      V.push_back(i + 1);
      second.clear();
      continue;
    }
    second.insert(inp[i]);
  }
  if (res == 0) {
    printf("-1");
    return 0;
  }
  if (V.back() != n)
    ;
  { V.back() = n; }
  printf("%d\n", res);
  for (i = 0; i < V.size() - 1; i++) {
    printf("%d %d\n", V[i] + 1, V[i + 1]);
  }
}