Codeforce915C

C. Permute Digits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integer numbers a and b. Permute (change order) of the digits of a to construct maximal number not exceeding b. No number in input and/or output can start with the digit 0.

It is allowed to leave a as it is.

Input

The first line contains integer a (1 ≤ a ≤ 10¹⁸). The second line contains integer b (1 ≤ b ≤ 10¹⁸). Numbers don't have leading zeroes. It is guaranteed that answer exists.

Output

Print the maximum possible number that is a permutation of digits of a and is not greater than b. The answer can't have any leading zeroes. It is guaranteed that the answer exists.

The number in the output should have exactly the same length as number a. It should be a permutation of digits of a.

Examples

Input

123
222

Output

213

Input

3921
10000

Output

9321

Input

4940
5000

Output

4940

题意：给两个数a和b,可以打乱a每位数的顺序组成一个新的数c,
求满足c<=b的最大,保证结果一定存在。

分析：先求出a的每位数字，然后排个序，再dfs。

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int A[100],B[100],C[100],alen,blen;
int vis[30],ans[30],K=-1;
int cmp(int x,int y)
{return x>y;}

int f(int a[],long long x)
{
    int len=0;
    while(x)
    {
        a[len++]=x%10;
        x/=10;
    }
    return len;
}

int dfs(int k)//k为当前构造到第k位数 
{
    if(k==alen) return 1;
    int last=-1;//前一个数 
    for(int i=0;i<alen;i++)
    {    
        if(last==A[i]) continue;
        if(vis[i]==0&&A[i]<=B[k])
        {
            if(k==0&&A[i]==0) continue;
            vis[i]=1;
            ans[k]=A[i];
            last=A[i];
            if(A[i]<B[k]) 
            {
                K=k;
                return 1;
            }
            if(dfs(k+1)) return 1;
            vis[i]=0;
        }
    }
    return 0;
}

int main()
{
    long long a,b;
    scanf("%lld%lld",&a,&b);
    alen=f(A,a);
    blen=f(C,b);
    for(int i=0;i<blen;i++)
    B[i]=C[blen-i-1]; 
    sort(A,A+alen,cmp);//排序 
    if(alen<blen)
    {
        for(int i=0;i<alen;i++)
            printf("%d",A[i]);
        printf("
");
    }
    else//alen==blen
    {
        dfs(0);
        if(K>=0)
        {
            for(int i=0;i<=K;i++)
            printf("%d",ans[i]);
            for(int i=0;i<alen;i++)
            if(!vis[i]) printf("%d",A[i]);
        }
        else
        {
            for(int i=0;i<alen;i++)
            printf("%d",ans[i]);
        }
        printf("
");
    }
    return 0;
}