dfs Gym

AbdelKader enjoys math. He feels very frustrated whenever he sees an incorrect equation and so he tries to make it correct as quickly as possible!

Given an equation of the form: A₁ o A₂ o A₃ o ... o A_n  =  0, where o is either + or -. Your task is to help AbdelKader find the minimum number of changes to the operators + and -, such that the equation becomes correct.

You are allowed to replace any number of pluses with minuses, and any number of minuses with pluses.

Input

The first line of input contains an integer N (2 ≤ N ≤ 20), the number of terms in the equation.

The second line contains N integers separated by a plus + or a minus -, each value is between 1 and 10⁸.

Values and operators are separated by a single space.

Output

If it is impossible to make the equation correct by replacing operators, print  - 1, otherwise print the minimum number of needed changes.

Examples

Input

7
1 + 1 - 4 - 4 - 4 - 2 - 2

Output

3

Input

3
5 + 3 - 7

Output

-1



题目翻译：一串数字改变其的符号，让sum为0，输出最小的改变次数，不存在则输出-1


运用算法DFS


ac代码：

#include<iostream>
using namespace std;
int a[25],n,ans;
void dfs(int index,int sum,int c)
{
　　if (index==n+1){
　　　　if (sum==0){
　　　　　　ans=ans<c?ans:c;
　　　　}
　　return ;
　　}
　　int j;
　　for (j=0;j<2;j++){ 　　　　 //只存在 + 或者是 - 两种情况
　　　　if (j==0){
　　　　　　if (a[index]<0)
　　　　　　　　dfs(index+1,sum-a[index],c+1);
　　　　else
　　　　　　dfs(index+1,sum+a[index],c);
　　　　}
　　　　else if (a[index]<0)
　　　　　　dfs(index+1,sum+a[index],c);
　　　　else
　　　　　　dfs(index+1,sum-a[index],c+1);
　　}
return ;

}
int main()
{
　　//scanf("%d",&n);
　　char op;
　　int i,j;
　　ans=99999999;
　　cin>>n;
　　for (i=1;i<=n;i++){
　　　　if (i==1)
　　　　　　scanf("%d",&a[i]);
　　　　else{
　　　　　　scanf(" %c %d",&op,&a[i]);
　　　　　　if (op=='-')
　　　　　　　　a[i]=-a[i];
　　　　　}
　　}
　　dfs(2,a[1],0);
　　　if (ans==99999999)
　　　　cout<<"-1 ";
　　else
　　　　cout<<ans<<endl;
　　return 0;
}