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;
long long big = 1000000007ll;
int l, n, m, q, k;
std::vector<int> A;
std::vector<int> num;
int main() {
  int a, b, c;
  long long c1, c2, c3, c4, c5;
  int t;
  scanf("%d %d", &n, &k);
  for (c1 = 0; c1 < n; c1++) {
    scanf("%d", &a);
    A.push_back(a);
  }
  for (c1 = 0; c1 < 1100000; c1++) {
    num.push_back(0);
  }
  int ans = 0;
  int nu = 0;
  int l = 0;
  a = 0;
  b = 0;
  int a1 = 0;
  int a2 = 0;
  ;
  while (b < n) {
    num[A[b]]++;
    if (num[A[b]] == 1) {
      nu++;
      l++;
    }
    while (nu > k) {
      num[A[a]]--;
      if (num[A[a]] == 0) {
        nu--;
      }
      a++;
      l--;
    }
    if ((b - a) > ans) {
      ans = b - a;
      a1 = a;
      a2 = b;
    }
    b++;
    l++;
  }
  printf("%d %d\n", a1 + 1, a2 + 1);
  return 0;
}
