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

41 3 2 3

Output

First

Input

22 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.

水题，不要想复杂了。一开始是先取奇数，所以如果sum是奇数，就可以直接取完，如果sum是偶数的话，我们看看里面是不是有奇数，如果有奇数的话，肯定就是奇数赢了，因为剩下的是奇数，另外一个人只能是偶数，所以剩下的还是奇数，所以这样的话，就是奇数的赢。但是，如果里面没有奇数，就是偶数赢了。

#include <iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<map>#include<string.h>#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;const int maxn=1e6+10;int num[maxn];int main(){   int n;   scanf("%d",&n);   int flag_odd=0;   ll sum=0;   for(int i=1;i<=n;i++)   {       scanf("%d",&num[i]);       sum += num[i];       if(num[i]%2!=0)        flag_odd=1;   }   if(sum%2==1||flag_odd)    printf("First\n");   else     printf("Second\n");   return 0;}