Drazil is playing a math game with Varda.

Let‘s define  for positive integer x as a product of factorials of its digits. For example, .

First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:

1. x doesn‘t contain neither digit 0 nor digit 1.

2.  = .

Help friends find such number.

The first line contains an integer n (1 ≤ n ≤ 15) — the number of digits in a.

The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes.

41234

33222

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() {    if(Head == Tail) {        int l = fread(buffer, 1, BufferSize, stdin);        Tail = (Head = buffer) + l;    }    return *Head++;}int read() {    int x = 0, f = 1; char c = getchar();    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }    return x * f;}#define maxn 6000010#define maxk 20int n, sum, fact[maxk], F[maxn], f[maxn][11];char num[maxk];bool vis[maxn];bool Less(int* a, int* b) {	for(int i = 9; i >= 2; i--) if(a[i] != b[i])		return a[i] < b[i];	return 0;}int main() {	n = read();	scanf("%s", num + 1);		fact[0] = 1;	for(int i = 1; i <= 9; i++) fact[i] = fact[i-1] * i;	sum = 1;	for(int i = 1; i <= n; i++) sum *= fact[num[i]-‘0‘];	vis[1] = 1;	for(int i = 1; i <= sum; i++) if(vis[i]) {//		printf("%d ", i);		for(int k = 2; k <= 9 && fact[k] * i <= sum; k++) {			int t = fact[k] * i;//			printf("[%d]", t);			vis[t] = 1;			if(F[t] < F[i] + 1) {				F[t] = F[i] + 1;				memcpy(f[t], f[i], sizeof(f[i]));				f[t][k]++;			}			if(F[t] > F[i] + 1) continue;			f[i][k]++;			if(Less(f[t], f[i])) memcpy(f[t], f[i], sizeof(f[i]));			f[i][k]--;		}	}		for(int i = 9; i >= 2; i--)		for(int j = 1; j <= f[sum][i]; j++)			putchar(i + ‘0‘);	putchar(‘\n‘);		return 0;}

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <stack>#include <vector>#include <queue>#include <cstring>#include <string>#include <map>#include <set>using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() {    if(Head == Tail) {        int l = fread(buffer, 1, BufferSize, stdin);        Tail = (Head = buffer) + l;    }    return *Head++;}int read() {    int x = 0, f = 1; char c = getchar();    while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }    while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }    return x * f;}#define maxn 20int n, cnt[maxn];char num[maxn];int main() {	n = read();	scanf("%s", num + 1);		for(int i = 1; i <= n; i++) {		if(num[i] == ‘9‘) {			cnt[7]++; cnt[3]++; cnt[3]++; cnt[2]++;		}		if(num[i] == ‘8‘) {			cnt[7]++; cnt[2]++; cnt[2]++; cnt[2]++;		}		if(num[i] == ‘7‘) {			cnt[7]++;		}		if(num[i] == ‘6‘) {			cnt[5]++; cnt[3]++;		}		if(num[i] == ‘5‘) {			cnt[5]++;		}		if(num[i] == ‘4‘) {			cnt[3]++; cnt[2]++; cnt[2]++;		}		if(num[i] == ‘3‘) {			cnt[3]++;		}		if(num[i] == ‘2‘) {			cnt[2]++;		}	}		for(int i = 7; i >= 2; i--)		for(int j = 1; j <= cnt[i]; j++) putchar(i + ‘0‘);	putchar(‘\n‘);		return 0;}