Codeforce Round #221 Div2 C

C. Divisible by Seven

time limit per test 1 second

memory limit per test 256 megabytes

 

You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.

Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.

Input

The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 10⁶ characters.

Output

Print a number in the decimal notation without leading zeroes — the result of the permutation.

If it is impossible to rearrange the digits of the number a in the required manner, print 0.

Sample test(s)

Input

1689

Output

1869

Input

18906

Output

18690

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll long long
#define maxn 1000005
#define pi acos(-1.0)
int n, m;
char s[maxn];
int a[maxn];
bool cmp(int a, int b){ return a > b; }
int main(){
    scanf("%s", s);
    n = strlen(s);
    memset(a, -1, sizeof a);
    sort(s, s + n,cmp);
    if (s[4] == '0'||n==4){
        printf("1869");
        for (int i = 0; i < n - 4; i++)printf("0");
        return 0;
    }
    for (int i = 0; i < n; i++){
        if (s[i] == '1'&&a[0] == -1)a[0] = i;
        if (s[i] == '6'&&a[1] == -1)a[1] = i;
        if (s[i] == '8'&&a[2] == -1)a[2] = i;
        if (s[i] == '9'&&a[3] == -1)a[3] = i;
    }
    int x = 0;
    for (int i = 0; i < n; i++){
        if (i == a[0] || i == a[1] || i == a[2] || i == a[3])continue;
        printf("%c", s[i]);
        x = x * 10 + (s[i] - '0');
        x = x % 7;
    }
    int t = x;
    for (int i = 0; i < 4; i++)
    for (int j = 0; j < 4; j++)
    for (int x = 0; x < 4; x++)
    for (int y = 0; y < 4; y++)
        if (i != j&&i != x&&i != y&&j != x&&j != y&&x != y){
            if ((t * 10000 + (s[a[i]] - '0') * 1000 + (s[a[j]] - '0') * 100 + (s[a[x]] - '0') * 10 + (s[a[y]] - '0')) % 7 == 0){
                printf("%c%c%c%c", s[a[i]], s[a[j]], s[a[x]], s[a[y]]);
                return 0;
            }
        }
}