You have an array a[1], a[2], ..., a[n], containing distinct integers from 1 to n. Your task is to sort this array in increasing order with the following operation (you may need to apply it multiple times):

  * choose two indexes, i and j (1 ≤ i < j ≤ n; (j - i + 1) is a prime number); 
  * swap the elements on positions i and j; in other words, you are allowed to apply the following sequence of assignments: tmp = a[i], a[i] = a[j], a[j] = tmp (tmp is a temporary variable). 



You do not need to minimize the number of used operations. However, you need to make sure that there are at most 5n operations.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n distinct integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).

Output

In the first line, print integer k (0 ≤ k ≤ 5n) — the number of used operations. Next, print the operations. Each operation must be printed as "i j" (1 ≤ i < j ≤ n; (j - i + 1) is a prime).

If there are multiple answers, you can print any of them.

Examples

Input

3
3 2 1


Output

1
1 3


Input

2
1 2


Output

0


Input

4
4 2 3 1


Output

3
2 4
1 2
2 4
#include <bits/stdc++.h>
using namespace std;
bool notprime[100000];
int primes[100000];
int sz = 0;
void sieve(void) {
  for (long long i = 2; i < 100000; i++) {
    if (!notprime[i]) {
      primes[sz++] = i;
      for (long long j = i * i; j < 100000; j += i) {
        notprime[j] = 1;
      }
    }
  }
}
int v[100000], a[100000], b[100000];
int main(void) {
  int n, i, x;
  sieve();
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    x--;
    v[i] = x;
    a[x] = i;
    b[i] = x;
  }
  sort(v, v + n);
  vector<pair<int, int> > ans;
  for (i = 0; i < n; i++) {
    int currind = a[v[i]];
    int currval = v[i];
    while (currind != i) {
      int j = upper_bound(primes, primes + sz, currind - i + 1) - primes;
      j--;
      int nextind = currind - primes[j] + 1;
      int nextval = b[nextind];
      a[nextval] = currind;
      a[currval] = nextind;
      b[currind] = nextval;
      b[nextind] = currval;
      ans.push_back(make_pair(currind, nextind));
      currind = nextind;
    }
  }
  printf("%d\n", ans.size());
  for (i = 0; i < ans.size(); i++) {
    printf("%d %d\n", ans[i].second + 1, ans[i].first + 1);
  }
  return 0;
}
