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 N, K;
int A[500005];
int CNT[1000005];
int DiffNums;
void Add(int i) {
  if (CNT[A[i]] == 0) DiffNums++;
  CNT[A[i]]++;
}
void Remove(int i) {
  if (CNT[A[i]] == 1) DiffNums--;
  CNT[A[i]]--;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; i++) scanf("%d", &A[i]);
  int ST = 0, EN = 0;
  int MaxLen = 0;
  int Left, Right;
  while (EN < N) {
    while (DiffNums <= K && EN < N) Add(EN++);
    if (EN - ST - (DiffNums > K ? 1 : 0) > MaxLen) {
      MaxLen = EN - ST - (DiffNums > K ? 1 : 0);
      Left = ST + 1;
      Right = EN - (DiffNums > K ? 1 : 0);
    }
    while (DiffNums > K) Remove(ST++);
  }
  printf("%d %d\n", Left, Right);
}
