题干：

You've got an n × m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.

You are really curious about prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors: itself and number one. For example, numbers 2, 3, 5 are prime and numbers 1, 4, 6 are not.

A matrix is prime if at least one of the two following conditions fulfills:

• the matrix has a row with prime numbers only;
• the matrix has a column with prime numbers only;

Your task is to count the minimum number of moves needed to get a prime matrix from the one you've got.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500) — the number of rows and columns in the matrix, correspondingly.

Each of the following n lines contains m integers — the initial matrix. All matrix elements are positive integers. All numbers in the initial matrix do not exceed 105.

The numbers in the lines are separated by single spaces.

Output

Print a single integer — the minimum number of moves needed to get a prime matrix from the one you've got. If you've got a prime matrix, print 0.

Examples

Input

3 3
1 2 3
5 6 1
4 4 1

Output

1

Input

2 3
4 8 8
9 2 9

Output

3

Input

2 2
1 3
4 2

Output

0

Note

In the first sample you need to increase number 1 in cell (1, 1). Thus, the first row will consist of prime numbers: 2, 2, 3.

In the second sample you need to increase number 8 in cell (1, 2) three times. Thus, the second column will consist of prime numbers: 11, 2.

In the third sample you don't have to do anything as the second column already consists of prime numbers: 3, 2.

 

题目大意：

   给出定义：素数矩阵是指一个矩阵中存在至少一行和一列全是素数的矩阵。现在给你一个矩阵,你可以选择矩阵中任意一个元素加1,问最少需要多少次这样的操作才能把这个矩阵变成一个素数矩阵。

解题报告：

  这题一眼就是一个n^2logn的写法，，，但是写这篇题解的时候发现其实是可以n^2的，，但是因为给了个2s所以时间还算宽裕就直接n^2logn了。。

   言归正传，其实就是个素数打表然后二分预处理出每个数对应的可以变成的素数的值，同时维护一个求个差的最小值就好了。。说到n^2其实也不难想，，就是把二分的过程给优化掉，，因为预处理的时候就是单调的所以不需要二分查找了做了一些无用的操作。。直接数组递推过去就可以了、。（这一招很常用啊，虽然在这一题中不明显但是有的时候就需要正着递推一遍反着递推一遍，线性就可以得到我们想要的东西，，比如还是那个经典题Fountain）

  另外啊这题因为数据量1e5所以打表就用了简单的nlogn，，懒得写线性筛了、、、

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e6 + 5;
int su[MAX],cnt;
bool isprime[MAX];
int a[550][550];
int qq[550][550];
void prime() {memset(isprime,1,sizeof isprime);isprime[1]=isprime[0]=0;for(int i = 2; i<=MAX; i++) {if(isprime[i]) {su[++cnt] = i;for(int j = 2*i; j<=MAX; j+=i) isprime[j] = 0;}}
}
int main()
{prime();int n,m;cin>>n>>m;for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {scanf("%d",&a[i][j]);}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {int pos = lower_bound(su+1,su+cnt+1,a[i][j]) - su;qq[i][j] = su[pos];}}ll minn = 0x3f3f3f3f3f;for(int i = 1; i<=n; i++) {ll tmp = 0;for(int j = 1; j<=m; j++) tmp += qq[i][j] - a[i][j];minn = min(minn,tmp);}for(int j = 1; j<=m; j++) {ll tmp = 0;for(int i = 1; i<=n; i++) tmp += qq[i][j] - a[i][j];minn = min(minn,tmp);}printf("%lld\n",minn);return 0 ;}