Codeforces Round #429 (Div. 2) 841B Godsend(签到题)

B. Godsend

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

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依旧签到题……emmmm……谁会赢呢……第一个玩家只能取和为奇数的一组数，第二个玩家只能取和为偶数的，

①如果总和为奇数，肯定就First赢，毫无悬念

②如果总和为偶数，a.不包括奇数，全部是偶数的话，那么第一个玩家就没有任何可以取的机会了，Second赢

                              b.包括n个奇数，第一个玩家只需要取走所有的偶数外加n-1个奇数，那么第一个玩家取完以后只剩下1个奇数，First赢

（orz这种题也WA了两次额，第一次想的不够全面，第二次……竟然是因为数组开小了……敲黑板啊！数组开小RE这种掉分简直不能接受）

#include<iostream>   
using namespace std;
typedef long long ll;
ll a;
int n, k,sum,count1,count2;
int b[1000005];
int main()
{
    while (cin >> n )
    {
        sum = 0;
        count1 = 0; count2 = 0;
        for (int i = 0; i < n; i++)
        {
            cin >> a;
            b[i] = a % 2;
            sum += b[i];
            if (b[i] % 2) count1++;
            else count2++;
        }
        if (sum % 2) cout << "First" << endl;
        else {
            if (count1 > 0) cout << "First" << endl;
            else cout << "Second" << endl;
        }
    }
    return 0;
}