Valera has array a, consisting of n integers a0,a1,...,an-1, and function f(x), taking an integer from 0 to 2n-1 as its single argument. Value f(x) is calculated by formula 
, where value bit(i) equals one if the binary representation of number xcontains a 1 on the i-th position, and zero otherwise.
For example, if n=4 and x=11 (11=20+21+23), then f(x)=a0+a1+a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0≤x≤m.
Input
The first line contains integer n (1≤n≤105) — the number of array elements. The next line contains n space-separated integersa0,a1,...,an-1 (0≤ai≤104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn-1 — the binary representation of number m. Numberm equals .
Output
Print a single integer — the maximum value of function f(x) for all .
Sample test(s)
input
2
3 8
10
output
3
input
5
17 0 10 2 1
11010
output
27
Note
In the first test case m=20=1,f(0)=0,f(1)=a0=3.
In the second sample m=20+21+23=11, the maximum value of function equals f(5)=a0+a2=17+10=27.
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思路：从左往右读，读到0就积累，读到1的时候，就看是修改这个1为0然后把前面的0改为1更大，还是不变更大。
修改的话，就从当前这个修改成0的1开始积累。
最终修改结果就是我们想要的x。
完整代码：
/*62ms,500KB*/
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxm = 100005;
int a[maxm];
char s[maxm];
int main()
{
int n, sum, maxn, i;
scanf("%d", &n);
for (i = 0; i < n; ++i)
scanf("%d", &a[i]);
getchar();
gets(s);
sum = maxn = 0;
for (i = 0; i < n; ++i)
{
if (s[i] & 15)
{
///看是前面的一串0大，还是当前位置的1大
if (maxn > a[i])
{
sum += maxn;
maxn = a[i]; ///重新积累
}
else sum += a[i];
}
else maxn += a[i];///积累maxn
}
printf("%d\n", sum);
return 0;
}
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