Description
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 ≤ 1000000) — length of the array.
Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 1000000000).
Output
Output answer in single line.
"First", if first player wins, and "Second" otherwise (without quotes).
Sample Input
4
1 3 2 3
Sample Output
First
题意：给你n个数，第一个人能删去一个连续的子串当且只当这个连续子串所有数之和为奇数，同理第二个人删的是和为偶数的子串，谁不能操作了谁输。考虑这个总串和为偶数的情况，因为为奇数的时候第一个人直接全删了。
当总和为偶数的时候，如果所有的数都是偶数的话，第一个人就没有办法删了，所以第二个人赢，不然那第一个先删去一段子串，剩余的子串之和肯定为奇数。第二个人删去偶数，肯定不能全删了，剩余的肯定是奇数，
那么第一个人必赢。综上，只有所有的数都是偶数的时候第二个人才赢，否则第一个人赢 。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <math.h>
#include <set>
using namespace std;
const int maxn=230;
int n,x,temp;
string s;
int main()
{
cin>>n;
bool flag=false;
while(n--)
{
scanf("%d",&x);
if(x%2)
flag=true;
}
if(flag==false)
cout<<"Second"<<endl;
else
cout<<"First"<<endl;
return 0;
}
/**********************************************************************
Problem: 2220
User: therang
Language: C++
Result: AC
Time:256 ms
Memory:2020 kb
**********************************************************************/