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 MAXN = 1000005;
int occ[MAXN], a[MAXN];
int n, k, l, sol, cnt, lo, hi;
void move(int r) {
  for (; l <= r; l++) {
    if (occ[a[l]] == l) {
      occ[a[l]] = -1;
      break;
    }
  }
  l++;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  memset(occ, -1, sizeof occ);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (occ[a[i]] == -1) cnt++;
    occ[a[i]] = i;
    if (cnt <= k) {
      if (i - l + 1 > sol) {
        lo = l;
        hi = i;
        sol = i - l + 1;
      }
      continue;
    }
    move(i);
    cnt--;
    if (i - l + 1 > sol) {
      lo = l;
      hi = i;
      sol = i - l + 1;
    }
  }
  cout << lo + 1 << ' ' << hi + 1 << endl;
  return 0;
}
