## solution

OEIS: https://oeis.org/A002187。愚直にgrundy数を求めて周期性。$O(n \log n)$。

## note

yukicoderでこんな問題が出たらほぼ間違いなく実験して周期性かOEISでしょという経験則がある。 まあゆるふわなのでOK。

## implementation

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define ALL(x) begin(x), end(x)
using namespace std;

/**
* @sa https://oeis.org/A002187
* @note > Has period 34 with the only exceptions at n=0, 14, 16, 17, 31, 34 and 51.
*/
vector<int> make_grundy(int k) {
vector<int> grundy(k);
grundy[0] = 0;
REP3 (len, 1, k) {
vector<int> used;
used.push_back(grundy[max(0, len - 2)]);
REP3 (l, 3, len + 1) {
used.push_back(grundy[l - 3] ^ grundy[len - l]);
}
sort(ALL(used));
used.erase(unique(ALL(used)), used.end());
int g = 0;
while (g < used.size() and g == used[g]) {
++ g;
}
grundy[len] = g;
}
return grundy;
}

bool solve(int n, vector<int> a) {
vector<int> table = make_grundy(34 * 3);
auto grundy = [&](int k) { return table[k < 34 * 3 ? k : k % 34 + 34 * 2]; };

sort(ALL(a));
a.erase(unique(ALL(a)), a.end());

int sum_g = 0;
for (int l = 0; l < a.size(); ) {
int r = l + 1;
while (r < a.size() and a[r] - a[l] == r - l) ++ r;
sum_g ^= grundy(r - l);
l = r;
}
return sum_g;
}

int main() {
int n; cin >> n;
vector<int> a(n);
REP (i, n) cin >> a[i];

bool ans = solve(n, a);

cout << (ans ? "First" : "Second") << endl;
return 0;
}