Codeforces Round #429 (Div. 2) B. Godsend

B. Godsend

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

Input

First line of input data contains single integer n (1 ≤ n ≤ 10⁶) — length of the array.

Next line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹).

Output

Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).

Examples

Input

4 1 3 2 3

Output

First

Input

2 2 2

Output

Second

Note

In first sample first player remove whole array in one move and win.

In second sample first player can't make a move and lose

 

题意：给一段数字，第一个人能拿走和为奇数的一段子串，第二个人能拿走和为偶数的一段字串。当一个人没有子串可拿时便输掉了比赛。问谁赢了比赛。

题解：如果开始一段数字和为奇数则第一个人全部拿完，第二个人没有可拿的了，第一个人赢。如果开始和为偶数且不存在奇数则第一个人没有可拿的子串输掉比赛。如果开始和为偶数且存在奇数则第一个人必胜。

 

代码：

#include <cstdio>
int main(){
      int n,sum=0,x;
      int flag=0;
      scanf("%d",&n);
      for(int i=0;i<n;i++){
            scanf("%d",&x);
            sum+=x;
            if(x%2!=0) flag=1;
       }
      if(sum%2!=0) printf("First\n");
      else{
             if(flag) printf("First\n");
            else printf("Second\n");
       }
      return 0;
}