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;
double Fib(double x, double z);
string BinDec(unsigned long long x);
string StringInt(long long x);
unsigned long long StringInt(string x);
unsigned long long BinDec(string x);
unsigned long long POWMOD(unsigned long long x, unsigned long long y,
                          unsigned long long mod);
unsigned long long POWE(long long, long long);
long long fast_pow(long long a, long long b, long long mod) {
  if (b == 0) return 1ll;
  long long ans = 1;
  if (b % 2) return ans = (fast_pow(a, b - 1, mod) * a) % mod;
  ans = (fast_pow(a, b / 2, mod) % mod);
  return ((ans * ans) % mod);
}
int tree[100000 * 4] = {}, arr[100200] = {}, x;
void build(int in, int s, int e) {
  if (s == e) {
    tree[in] = s;
    return;
  }
  int m = (s + e) / 2;
  build(in * 2, s, m);
  build(in * 2 + 1, m + 1, e);
  if (arr[tree[in * 2]] > arr[tree[in * 2 + 1]])
    tree[in] = tree[in * 2 + 1];
  else
    tree[in] = tree[in * 2];
}
void up(int in, int s, int e, int ind, int v) {
  if (s > ind || ind > e) return;
  if (s == e && s == ind) {
    arr[s] = v;
    tree[in] = s;
    return;
  }
  int m = (s + e) / 2;
  up(in * 2, s, m, ind, v);
  up(in * 2 + 1, m + 1, e, ind, v);
  if (arr[tree[in * 2]] > arr[tree[in * 2 + 1]])
    tree[in] = tree[in * 2 + 1];
  else
    tree[in] = tree[in * 2];
}
long long qur(int in, int s, int e, int f, int l) {
  if (s > e || s > l || f > e) return 100001;
  if (f <= s && l >= e) return tree[in];
  int m = (s + e) / 2;
  long long v, i = qur(in * 2, s, m, f, l), j = qur(in * 2 + 1, m + 1, e, f, l);
  if (arr[i] > arr[j])
    v = j;
  else
    v = i;
  return v;
}
bool p[100210] = {};
int S = 0, SS = 0;
vector<int> v;
vector<pair<int, int> > ans;
pair<int, int> vv[100010] = {};
void sort() {
  int t;
  for (int i = 0; i < x; i++) {
    int j = qur(1, 0, x - 1, i, x - 1);
    if (i == j) continue;
    int k = j - i + 1;
    if (!p[k]) {
      ans.push_back({i, j});
      t = arr[i];
      up(1, 0, x - 1, i, arr[j]);
      up(1, 0, x - 1, j, t);
    } else if (k % 2 == 0) {
      if (p[j - i - vv[k].second + 1]) S++;
      ans.push_back({i + vv[k].second, j});
      t = arr[i + vv[k].second];
      up(1, 0, x - 1, i + vv[k].second, arr[j]);
      up(1, 0, x - 1, j, t);
      if (p[i + vv[k].second + -i - 1 + 1]) S++;
      ans.push_back({i + 1, i + vv[k].second});
      t = arr[i + 1];
      up(1, 0, x - 1, i + 1, arr[i + vv[k].second]);
      up(1, 0, x - 1, i + vv[k].second, t);
      if (p[i + 1 - i + 1]) S++;
      ans.push_back({i, i + 1});
      t = arr[i + 1];
      up(1, 0, x - 1, i + 1, arr[i]);
      up(1, 0, x - 1, i, t);
    } else {
      k++;
      i--;
      if (p[j - i - vv[k].second + 1]) S++;
      ans.push_back({i + vv[k].second, j});
      t = arr[i + vv[k].second];
      up(1, 0, x - 1, i + vv[k].second, arr[j]);
      up(1, 0, x - 1, j, t);
      if (p[i + vv[k].second + -i - 1 + 1]) SS++;
      ans.push_back({i + 1, i + vv[k].second});
      t = arr[i + 1];
      up(1, 0, x - 1, i + 1, arr[i + vv[k].second]);
      up(1, 0, x - 1, i + vv[k].second, t);
      i++;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (long long i = 2; i <= 100000; i++) {
    if (!p[i]) {
      v.push_back(i);
      for (long long j = i * i; j <= 100000; j += i) p[j] = 1;
    }
  }
  for (int i = 4; i <= 100000; i += 2) {
    for (int j = 0; j < v.size(); j++) {
      if (!p[i - v[j]]) {
        vv[i] = {i - v[j], v[j]};
        break;
      }
    }
  }
  arr[100001] = 1e10;
  cin >> x;
  for (int i = 0; i < x; i++) cin >> arr[i];
  build(1, 0, x - 1);
  int j = 0;
  sort();
  if (S != 0 || SS != 0) cout << S << ' ' << SS << endl;
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << '\n';
  ;
  return 0;
}
