CodeForces-915C Permute Digits

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

input

3921
10000

output

input

4940
5000

output4940 题目链接


题意 
给两个数a和b，可以任意调换a中数字的位置，问由a变成的最大的不超过b的数是什么？

分析
暴搜加剪枝。因为结果一定存在，那么可以分为两种情况，lena<lenb或是lena==lenb。当lena<lenb时，从大到小输出a中的数即可。
当lena==lenb时，我们尽量保持和b的对应位相等，若一出现某位的数小于b上的数，剩下的数从大到小输出就好。
但是，会遇到前面都匹配的是相等的数，到了这一位，没有小于等于b对应位的数，这样构造出来的值就会大于b了，
这样子是不行的，所以我们得回溯，这样就需要dfs了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
using namespace std;
typedef long long LL;
const int maxn = 400;
const int mod = 772002+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
//#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))

const int inf = 0x3f3f3f3f;
char a[20],b[20],ans[20];
int cnt[10];
int lena,lenb;
bool dfs(int idx,int flag){
    if(idx==lenb) return true;

    for(int ia=9;ia>=0;ia--){
        if(cnt[ia]){
            if(flag ||ia+'0'==b[idx]){//此前b已比ans大了
                ans[idx]=ia+'0';
                cnt[ia]--;
                if(dfs(idx+1,flag)) return true;
                cnt[ia]++;
            }else if(ia+'0'<b[idx]){
                ans[idx]=ia+'0';
                cnt[ia]--;
                if(dfs(idx+1,flag|1))
                    return true;
            }
        }
    }
    return false;
}
int main(){
    scanf("%s%s",a,b);
    lena=strlen(a);
    lenb=strlen(b);

    if(lena<lenb){
        sort(a,a+lena);
        for(int i=lena-1;i>=0;i--) putchar(a[i]);
    }else{
        ms(cnt,0);
        for(int i=0;i<lena;i++) cnt[a[i]-'0']++;
        dfs(0,0);
        ans[lena]=0;
        puts(ans);
    }
}