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

题目：

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 integersa₁, 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.

题意：给出一个数组，有两个人玩游戏，游戏规则：第一个人选取连续的子区间并且区间的和为奇数，第二个人选取连续的子区间并且区间和为偶数。当有人不能选择这样的一个区间时，那么这个人就输了，问那个人会赢？

思路：思维题。容易看出，当存在一个奇数的时候第一个人会赢，否则第二个人会赢。另奇数的个数为x，当x为偶数时，第一个人只需选取最后一个奇数前的区间，那么无论第二个人怎么选都会存在和为奇数的区间；当x为奇数时，此时整个数组的和为奇数满足条件，第一个人直接选择整个数组~

code：

#include<bits/stdc++.h>using namespace std;int a[1000005];int main(){        int n,i,x;        while(~scanf("%d",&n)){                x=0;                for(i=0;i<n;i++){                        scanf("%d",&a[i]);                        if(a[i]%2) x++;                }                if(x) puts("First");                else puts("Second");        }        return 0;}