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 mm[1000001];
int cc[1000001];
bool solve(int mid, int k, int a[], int n) {
  memset(mm, 0, sizeof(mm));
  int i, j;
  int cnt = 0;
  for (i = 1; i <= mid; i++) {
    mm[a[i]]++;
    if (mm[a[i]] == 1) cnt++;
  }
  i = 1;
  j = mid;
  while (j <= n) {
    if (cnt <= k) return true;
    mm[a[i]]--;
    if (mm[a[i]] == 0) cnt--;
    if (j + 1 <= n) {
      mm[a[j + 1]]++;
      if (mm[a[j + 1]] == 1) cnt++;
    }
    i++;
    j++;
  }
  return false;
}
int main() {
  int i, j, n, k, len, res = 0, ll, rr;
  scanf("%d %d", &n, &k);
  int a[n + 1];
  a[0] = -1;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  int l = 0, r = n + 1, mid;
  while (r - l > 1) {
    mid = (l + r) / 2;
    bool rr = solve(mid, k, a, n);
    if (rr == true)
      l = mid;
    else
      r = mid;
  }
  for (i = 1; i <= l; i++) {
    cc[a[i]]++;
    if (cc[a[i]] == 1) res++;
  }
  i = 1;
  j = l;
  while (j <= n) {
    if (res <= k) {
      printf("%d %d\n", i, j);
      break;
    }
    cc[a[i]]--;
    if (cc[a[i]] == 0) res--;
    if (j + 1 <= n) {
      cc[a[j + 1]]++;
      if (cc[a[j + 1]] == 1) res++;
    }
    i++;
    j++;
  }
  return 0;
}
