Permute Digits 915C

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

题解：

dfs,深搜，需要注意的是如果我们找到一个小于当前B中的值时，后面的只需要从小到大排列就好了

#include <bits/stdc++.h>
using namespace std;
char a[20],b[20];
int num[11],x[200],w[200],lb,la;
long long MAX=0,B;
inline long long MAx(long long z,long long w)
{
    if(z>w) return z;
    return w;
}
bool cmp(char a,char b)
{
    return a>b;
}
 
void dfs(long long ans,int len,int k,int v)
{
    if(len==la&&len<=lb&&ans<=B) {
        MAX=MAx(ans,MAX);
        ans=0;
        return;
    }
    for (int i = 9; i >=0 ; i--) {
        if (num[i])
        {
            if(v==0&&i==w[k])
            {
                num[i]--;
                ans=ans*10+i;
                dfs(ans,len+1,k+1,0);
                ans-=i;
                ans/=10;
                num[i]++;
            } else if(i<w[k])
            {
                num[i]--;
                ans=ans*10+i;
                for (int j = 9; j >=0 ; j--) {
                    for (int z = 0; z <num[j] ; ++z) {
                        ans=ans*10+j;
                    }
                }
                if(ans<B)
                {
                    MAX=MAx(MAX,ans);
                }
                ans=0;
            }
        }
    }
}
int main()
{
    scanf("%s",a);
    scanf("%s",b);
    la=(int)strlen(a);
    lb=(int)strlen(b);
    for (int i = 0; i <la ; ++i) {
        num[a[i]-'0']++;
        x[i]=a[i]-'0';
    }
    for (int i = 0;  i<lb; i++) {
        w[i]=b[i]-'0';
        B=B*10+w[i];
    }
    if(la<lb)
    {
        sort(a,a+la,cmp);
        for (int i = 0; i <la ; ++i) {
            printf("%c",a[i]);
        }
        printf("\n");
        return 0;
    }
    dfs(0,0,0,0);
    printf("%lld",MAX);
    return 0;
}