Problem: 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, ..., an (0 ≤ ai ≤ 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void dk_98() {}
bool is_prime(int n) {
  if (n == 1) {
    return false;
  }
  if (n == 2) {
    return true;
  }
  int factors = 0;
  for (int i = 2; i < n; i++) {
    if (n % i == 0) {
      factors++;
      break;
    }
  }
  if (factors == 0) {
    return true;
  }
  return false;
}
long long power(long long a, long long b) {
  long long temp = a;
  long long ans = 1;
  while (b >= 1) {
    if (b & 1) {
      ans *= temp;
      ans %= 1000000007;
    }
    b >>= 1LL;
    temp *= (temp % 1000000007);
    temp %= 1000000007;
  }
  return ans;
}
long long modFact(long long n, long long p) {
  if (n >= p) return 0;
  long long result = 1;
  for (long long i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long b, long long m) {
  long long x, y;
  long long g = gcdExtended(b, m, &x, &y);
  if (g != 1) return -1;
  return (x % m + m) % m;
}
long long modDivide(long long a, long long b, long long m) {
  a = a % m;
  long long inv = modInverse(b, m);
  return ((inv * a) % m);
}
long long npr(long long n, long long r) {
  return modDivide(modFact(n, 1000000007), modFact(n - r, 1000000007),
                   1000000007);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  dk_98();
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    int odd = 0, even = 0, sum = 0;
    for (int i = 0; i < (int)(n); i++) {
      cin >> arr[i];
      sum += arr[i];
      if (arr[i] & 1)
        odd++;
      else
        even++;
    }
    if (sum & 1) {
      cout << "First";
      return 0;
    } else {
      if (odd == 0) {
        cout << "Second";
        return 0;
      }
      if (odd & 1) {
        cout << "Second";
        return 0;
      }
      cout << "First";
    }
  }
}