There are two seagulls playing a very peculiar game. First they line up N unit squares in a line, all originally colored white.

Let L be the length of the longest continuous sub-segment of white unit squares. Let P be any prime that satisfies the condition that . The player can choose any P if it satisfies the conditions but if there is no value of P that does, then P is set to 1.

The current player then proceeds to choose any continuous sub-segment of white unit squares of length P and paint them black. The player who can’t make a move loses.

If both players play optimally and the first player starts, which player is going to win the game?

Input

The first line of input is T – the number of test cases.

Each test case contains a line with a single integer N (1 ≤ N ≤ 10⁷).

Output

For each test case, output on a single line "first" (without quotes) if the first player would win the game, or "second" if the second would win.

Example

2
2
5

second
first

题意：
每次可以取一段白色连续区间内p个，把白的染成黑色。不能染色的玩家失败。
整段区间内，最长的白色连续区间的长度为L,p是小于等于L/2的任意质数，如果没有质数，p可以取1；
思路：
手推，2,3是先手必败，1是先手必胜。
大于3的偶数，在中间取2，使留下来的两段长度相等。
大于3的奇数，在中间取3，使留下来的两段长度相等。
然后就像nim博弈只有两堆石子那样，跟着第二个人取，就可以坐等胜利了。

#include<cstdio>

int main()

{

    int T;

    scanf("%d",&T);

    while(T--){

        int n;

        scanf("%d",&n);

        if(n==||n==){printf("second\n");}

        else printf("first\n");

    }

    return ;

}