Problem Description

You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is si. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well.

However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?

Constraints

All input values are integers.1≤N≤1001≤si≤100

Input

Input is given from Standard Input in the following format:

N s1 s2 : sN

Output

Print the maximum value that can be displayed as your grade.

Example

Sample Input 1

3 5 10 15

Sample Output 1

25 Your grade will be 25 if the 10-point and 15-point questions are answered correctly and the 5-point question is not, and this grade will be displayed correctly. Your grade will become 30 if the 5-point question is also answered correctly, but this grade will be incorrectly displayed as 0.

Sample Input 2

3 10 10 15

Sample Output 2

35 Your grade will be 35 if all the questions are answered correctly, and this grade will be displayed correctly.

Sample Input 3

3 10 20 30

Sample Output 3

0 Regardless of whether each question is answered correctly or not, your grade will be a multiple of 10 and displayed as 0.

题意：给出 n 个数，可以选任意个数，使得这些数的和最大且不是 10 的倍数，输出这些数的和

思路：

首先对给出的 n 个数求和，若这些数的和不是 10 的倍数，则这 n 个数的和即为答案

如果这 n 个数的和是 10 的倍数，那么在这些数中找一个不是 10 的倍数的最小的数，然后用和减去他即可

在求和过程中，如果所有数都是 10 的倍数，那么答案为 0

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std; int main() { int n; scanf("%d",&n); int sum=0; int minn=INF; bool flag=false; for(int i=1; i<=n; i++){ int x; scanf("%d",&x); sum+=x; if(x!=0){ flag=true; minn=min(minn,x); } } if(!flag) printf("0\n"); else{ if(sum!=0) printf("%d\n",sum); else printf("%d\n",sum-minn); } return 0; }