Vanya is doing his maths homework. He has an expression of form ,
where x1, x2, ..., xn are
digits from 1 to 9, and
sign  represents
either a plus '+' or the multiplication sign '*'.
Vanya needs to add one pair of brackets in this expression so that to maximize the value of the resulting expression.

The first line contains expression s (1 ≤ |s| ≤ 5001, |s| is
odd), its odd positions only contain digits from 1 to 9,
and even positions only contain signs  +  and  * .

The number of signs  *  doesn't exceed 15.

In the first line print the maximum possible value of an expression.

Note to the first sample test. 3 + 5 * (7 + 8) * 4 = 303.

Note to the second sample test. (2 + 3) * 5 = 25.

Note to the third sample test. (3 * 4) * 5 = 60 (also many other variants are valid, for instance, (3) * 4 * 5 = 60).

/**

CF552E 字符串 表达式求值

题目大意：给定一个字符串仅仅是做1~9之间的加法和乘法，如今在表达式中加上一对括号。问怎样加才干使表达式的值最大

解题思路：左括号必须在一个*的后面，右括号必须在一个*的前面，假设不是这样一定不是最优。

有了这个结论，分成三部分算一下就能够了

*/

#include <string.h>

#include <iostream>

#include <algorithm>

#include <stdio.h>

using namespace std;

typedef long long LL;

char s[5005];

int n,a[105];

LL get(int l,int r)

{

    LL ans=0,tmp=1;

    for(int i=l;i<=r;i++)

    {

        if(isdigit(s[i]))

        {

            tmp*=s[i]-'0';

        }

        if(s[i]=='+')

        {

            ans+=tmp;

            tmp=1;

        }

    }

    ans+=tmp;

    tmp=1;

    LL sum=0;

    for(int i=0;i<n;i++)

    {

        if(i==l)

        {

            tmp*=ans;

            i=r;

        }

        else

        {

            if(isdigit(s[i]))

            {

                tmp*=s[i]-'0';

            }

            if(s[i]=='+')

            {

                sum+=tmp;

                tmp=1;

            }

        }

    }

    sum+=tmp,tmp=1;

   // printf(">>>%d %d %d %d\n",l,r,ans,sum);

    return sum;

}

int main()

{

    scanf("%s",s);

    n=strlen(s);

    int k=0;

    a[k++]=0;

    for(int i=0;i<n;i++)

    {

        if(s[i]=='*')

        {

            a[k++]=i-1;

            a[k++]=i+1;

        }

    }

    if(a[k-1]!=n-1)

        a[k++]=n-1;

    LL maxx=0;

    for(int i=0;i<k;i++)

    {

        for(int j=i;j<k;j++)

        {

            LL t=get(a[i],a[j]);

            maxx=max(maxx,t);

        }

    }

    printf("%lld\n",maxx);

    return 0;

}