Problem:
Score : 500 points

Problem Statement

There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i.

You and Lunlun the dachshund alternately perform the following operation (starting from you):

Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors.

The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?

Constraints

1 \leq N \leq 10^5

1 \leq a_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
a_2
:
a_N

Output

If you will win, print first; if Lunlun will win, print second.

Sample Input 1

2
1
2

Sample Output 1

first

Let Color 1 be red, and Color 2 be blue. In this case, the tree bears one red apple and two blue apples.

You should eat the red apple in your first turn. Lunlun is then forced to eat one of the blue apples, and you can win by eating the other in your next turn.

Note that you are also allowed to eat two apples in your first turn, one red and one blue (not a winning move, though).

Sample Input 2

3
100000
30000
20000

Sample Output 2

second

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

random_device(rd);
mt19937 rng(rd());
const long long FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();

struct custom_hash
{
	template<class T>
	unsigned long long operator()(T v) const
	{
		unsigned long long x = v;
		x += FIXED_RANDOM; x += 11400714819323198485ull;
		x = (x ^ (x >> 30)) * 13787848793156543929ull;
		x = (x ^ (x >> 27)) * 10723151780598845931ull;
		return x ^ (x >> 31);
	}
};

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U> using hash_table = gp_hash_table<T, U, custom_hash>;

template<class T>
void readi(T &x)
{
	x = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		x = x * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		x = -x;
	}
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int buf[20], n = 0;
	while(output)
	{
		buf[n] = ((output % 10));
		output /= 10;
		n++;
	}
	for (n--; n >= 0; n--)
	{
		putchar(buf[n] + '0');
	}
	return;
}
template<class T>
void ckmin(T &a, T b)
{
	a = min(a, b);
}
template<class T>
void ckmax(T &a, T b)
{
	a = max(a, b);
}
long long expo(long long a, long long e, long long mod)
{
	return ((e == 0) ? 1 : ((expo(a * a % mod, e >> 1, mod)) * ((e & 1) ? a : 1) % mod));
}
template<class T, class U>
void nmod(T &x, U mod)
{
	if (x >= mod) x -= mod;
}
template<class T>
T gcd(T a, T b)
{
	return (b ? gcd(b, a % b) : a);
}
template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define DBG(x) cerr << #x << " = " << x << endl
#define SZ(x) ((int) (x.size()))
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) x.begin(), x.end()

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-9;

#define MAGIC 347
#define SINF 10007
#define CO 1000007
#define INF 1000000007
#define BIG 1000000931
#define LARGE 1696969696967ll
#define GIANT 2564008813937411ll
#define LLINF 2696969696969696969ll
#define MAXN 100013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<pdd> vpd;

int N;
vector<ll> arr;
ll mx = 0;
bool one = false;

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	// cout << fixed << setprecision(10);
	// cerr << fixed << setprecision(10);
	// if (fopen("file.in", "r"))
	// {
	// 	freopen ("file.in", "r", stdin);
	// 	freopen ("file.out", "w", stdout);
	// }
	cin >> N;
	FOR(i, 0, N)
	{
		cin >> arr[i];
		if (arr[i] % 2 == 0)
		{
			cout << "first\n";
			return 0;
		}
	}
	cout << "second\n";
	//the winning player can make sure that exactly one is removed from each pile!
	//you can take 0 or 1 from each pile
	// cerr << "time elapsed = " << (clock() / (CLOCKS_PER_SEC / 1000)) << " ms" << endl;
	return 0;
}
/* READ READ READ
* int overflow, maxn too small, special cases (n=1?, two distinct?), cin.tie() interactive
* reread the problem, try small cases
* do smth instead of nothing
*/