C. Prime Swaps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

  HERE WE GO THROUGH A GREEDY STRATEGY AND ALWAYS BRING THE NUMBER TO ITS NEAREST POSITION THAT CAN BE DONE USING A VALID SWAP. IT IS ALWAYS CORRECT BECAUSE WE CAN SWAP TWO NEIGHBORING NUMBERS ALWAYS SO THIS WILL LEAD TO A SOLUTION.

            LOOK AT THE EASY CODE

#include<bits/stdc++.h>

using namespace std;

#define ll long long int 

bool pr[100010];

#define ff first

#define ss second

#define mp make_pair

#define pb push_back

vector<int> v;

int l;

void pre()

{

   for(int i=2;i<=100000;i++)

   {

      if(!pr[i])

      {

        v.pb(i);

        for(int j=2*i;j<=100000;j+=i)

        {

              pr[j]=1;

  }

  }

}

l=v.size();

}

int pos[100010];

int a[100010];

int main()

{

    pre();

     int n;

     cin>>n;

     for(int i=1;i<=n;i++)

     {

       cin>>a[i];

       pos[a[i]]=i;

 }

 vector<pair<int,int> > res;

 

 for(int i=1;i<=n;i++)

 {

        while(pos[i]!=i)

        {

          int lst=pos[i];

       

             int shiftm=abs(pos[i]-i)+1;

   vector<int> ::iterator it;

   it=upper_bound(v.begin(),v.end(),shiftm);

it--;

int val=*it;

// cout<<"value "<<val<<endl;

     int idx=pos[i]+1-val;

//      cout<<"idx "<<idx<<endl;

     res.pb(mp(pos[i],idx));

     

     pos[a[idx]]=pos[i];

     pos[i]=idx;

     

    // cout<<"sert pos of "<<i<<" as "<<pos[i]<<" and pos of "<<a[idx]<<" = "<<pos[a[idx]]<<endl;

     swap(a[idx],a[lst]);

}

 

}

int sz=res.size();

cout<<sz<<endl;

for(int i=0;i<sz;i++)

cout<<min(res[i].ff,res[i].ss)<<" "<<max(res[i].ff,res[i].ss)<<endl;

 return 0;

 }