Codeforces Round #104 (Div.2)补题报告

B - Lucky Mask

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.

Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.

Input

The only line contains two integers a and b (1 ≤ a, b ≤ 10⁵). It is guaranteed that number b is lucky.

Output

In the only line print a single number — the number c that is sought by Petya.

Examples

Input

1 7

Output

7

Input

100 47

Output

147

解题思路：从a开始遍历直到找到c，使c从左读到右的幸运数的组合数等于b，利用string的begin和end函数和reverse函数反转string.

ac代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
string mask(int a){
    string ret;
    while(a){
        if(a%10==4||a%10==7) {
            ret+=((a%10)+'0');
        }
        a/=10;
    }
    reverse(ret.begin(),ret.end());
    return ret;
}
int main()
{
    int a;
    string b;
    cin>>a>> b;
    int res= a + 1;
    while(mask(res)!= b){
        res++;
    }
    cout<<res<<endl;
    return 0;
}

D - Lucky Number 2

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya loves long lucky numbers very much. He is interested in the minimum lucky number d that meets some condition. Let cnt(x) be the number of occurrences of number x in number d as a substring. For example, if d = 747747, then cnt(4) = 2, cnt(7) = 4, cnt(47) = 2, cnt(74) = 2. Petya wants the following condition to fulfil simultaneously: cnt(4) = a₁, cnt(7) = a₂, cnt(47) = a₃, cnt(74) = a₄. Petya is not interested in the occurrences of other numbers. Help him cope with this task.

Input

The single line contains four integers a₁, a₂, a₃ and a₄ (1 ≤ a₁, a₂, a₃, a₄ ≤ 10⁶).

Output

On the single line print without leading zeroes the answer to the problem — the minimum lucky number d such, that cnt(4) = a₁, cnt(7) = a₂, cnt(47) = a₃, cnt(74) = a₄. If such number does not exist, print the single number "-1" (without the quotes).

Examples

Input

2 2 1 1

Output

4774

Input

4 7 3 1

Output

-1
解题思路：

如果c>d，第一个和最后一个分别是4和7，中间输出d个74，要使构成的数最小，4一定放在前面输出，7放在后面输出

如果c<d，第一个和最后一个分别是7和4，中间输出c个47，4一定在前面输出，7在最后一个4之前输出

如果c=d,如果a>c,则以4为开头和结尾输出，输出a-c-1个4，再输出c个47，b-c个7，再输出一个4

如果a=c，以7为开头和结尾输出，输出c个47，b-c-1个7

ac代码：

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
#define pf printf
#define sf scanf
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
int a,b,c,d;
 
int main()
{
    cin>>a>>b>>c>>d;
        if(abs(c-d)>1||a+b<=c+d||a<c||b<d||a<d||b<c)
        {
            puts("-1");
        }
        else if(c>d)
        {
            a=a-d,b=b-d;
            for1(i,a)pf("4");
            for1(i,d)pf("74");
            for1(i,b)pf("7");
            puts("");
        }
        else if(c==d)
        {
            if(a>c)
            {
                a-=1+c;
                b-=c;
                for1(i,a)pf("4");
                for1(i,c)pf("47");
                for1(i,b)pf("7");
                pf("4");
                puts("");
            }
            else
            {
                b-=1+c;
                pf("7");
                for1(i,c)pf("47");
                for1(i,b)pf("7");
                puts("");
            }
        }
        else
        {
            a-=c+1,b-=c+1;
            pf("7");
            for1(i,a)pf("4");
            for1(i,c)pf("47");
            for1(i,b)pf("7");
            pf("4");
            puts("");
        }
    return 0;
}