#include <bits/stdc++.h> 
using namespace std;  
 
typedef long long ll;  
typedef pair<int, int> ii;  
 
const int INF = 1e9;  
const ll LINF = 1e18;  
 
const int N = 5e2 + 5; 
const int MX = 1e7 + 5; 
 
int n, m; 
int a[N][N]; 
 
bool prime[MX]; 
 
void sieve() {
	for (int i = 2; i < MX; i++) prime[i] = true;   
 
	for (int i = 2; i * i < MX; i++) {
		if (prime[i]) {
			for (int j = i * i; j < MX; j += i) prime[j] = false;
		}
	}
}
 
// Gọi d(x) = khoảng cách của x đến số nguyên tố gần nhất mà >= x 
// Ta có nhận xét: d(x) <= ~230 với x <= 1e9, nên với 1 số x ta có thể trâu để tính d(x)
int d[N][N]; 
int sum_r[N]; // sum_r[i] = tổng d(x) của các số x có trong hàng i 
int sum_c[N]; // sum_c[j] = tổng d(x) của các số x có trong cột j
 
void precompute() {
	sieve(); 
 
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			while (!prime[a[i][j]]) {
				++a[i][j]; 
				++d[i][j]; 
			}
			sum_r[i] += d[i][j]; 
			sum_c[j] += d[i][j]; 
		}
	}
}
 
int main() {
	ios::sync_with_stdio(0); cin.tie(0);  	
	cin >> n >> m; 
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) cin >> a[i][j]; 
	}
 
	precompute();  
 
	int ans = INF; 
	for (int i = 1; i <= n; i++) ans = min(ans, sum_r[i]);  
	for (int j = 1; j <= m; j++) ans = min(ans, sum_c[j]); 
 
	cout << ans << '\n'; 
}

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3 3
1 2 3
4 9 8
2 3 9