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 MAX_N = (int)3e5 + 10;
int N;
int Ans[MAX_N][2], AnsP;
map<int, int> M;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    int x;
    scanf("%d", &x);
    if (M.find(x) == M.end()) {
      M[x] = i;
    } else {
      AnsP++;
      Ans[AnsP][0] = M[x];
      Ans[AnsP][1] = i;
      M.clear();
    }
  }
  if (AnsP == 0)
    printf("-1");
  else
    printf("%d", AnsP);
  Ans[0][1] = 0;
  for (int i = 1; i <= AnsP; i++) {
    if (Ans[i][1] != N && i == AnsP) Ans[i][1] = N;
    printf("\n%d %d", Ans[i - 1][1] + 1, Ans[i][1]);
  }
  return 0;
}
