CF1382B Sequential Nim

Description

There are $ n $ piles of stones, where the $ i $ -th pile has $ a_i $ stones. Two people play a game, where they take alternating turns removing stones. In a move, a player may remove a positive number of stones from the first non-empty pile (the pile with the minimal index, that has at least one stone). The first player who cannot make a move (because all piles are empty) loses the game. If both players play optimally, determine the winner of the game.

Input Format

The first line contains a single integer $ t $ ( $ 1\le t\le 1000 $ ) — the number of test cases. Next $ 2t $ lines contain descriptions of test cases. The first line of each test case contains a single integer $ n $ ( $ 1\le n\le 10^5 $ ) — the number of piles. The second line of each test case contains $ n $ integers $ a_1,\ldots,a_n $ ( $ 1\le a_i\le 10^9 $ ) — $ a_i $ is equal to the number of stones in the $ i $ -th pile. It is guaranteed that the sum of $ n $ for all test cases does not exceed $ 10^5 $ .

Output Format

For each test case, if the player who makes the first move will win, output "First". Otherwise, output "Second".

Explanation/Hint

In the first test case, the first player will win the game. His winning strategy is: 1. The first player should take the stones from the first pile. He will take $ 1 $ stone. The numbers of stones in piles will be $ [1, 5, 4] $ . 2. The second player should take the stones from the first pile. He will take $ 1 $ stone because he can't take any other number of stones. The numbers of stones in piles will be $ [0, 5, 4] $ . 3. The first player should take the stones from the second pile because the first pile is empty. He will take $ 4 $ stones. The numbers of stones in piles will be $ [0, 1, 4] $ . 4. The second player should take the stones from the second pile because the first pile is empty. He will take $ 1 $ stone because he can't take any other number of stones. The numbers of stones in piles will be $ [0, 0, 4] $ . 5. The first player should take the stones from the third pile because the first and second piles are empty. He will take $ 4 $ stones. The numbers of stones in piles will be $ [0, 0, 0] $ . 6. The second player will lose the game because all piles will be empty.