【CF Round 429 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.

【翻译】输入一个长度为n的序列，第一个人和第二个人轮流操作，第一个人先操作。第一个人可以取走和为奇数的连续一段，第二个人可以取走和为偶数的连续一段，但是不能不去。两人采取最优策略，最终谁会赢(第一个赢输出Fires,反之输出Second)。

题解：
         ①分类讨论就可以了：
               (1)如果整个序列和为奇数，那么开局First就胜利了。

               (2)如果整个序列和为偶数：
                     [1]如果这个序列没有奇数,那么First没法取，Second赢了。

                     [2]如果存在奇数，那么必会有偶数个奇数，那么First每次只消去一个奇数，到最后一定是他赢，因为Second始终没法消去奇数个奇数。

#include<stdio.h>
int main()
{
	int n,sum=0,_=0,a;scanf("%d",&n);
	while(n--)scanf("%d",&a),sum+=a,a&1?_=1:1;
	puts(sum&1||_?"First":"Second");return 0;
}//Paul_Guderian

祝母校60岁生日快乐！———Rice_Rabbit