Walk Alone has n×mntimes mn×m pieces of chocolates placed in a board with nnn rows and mmm columns The coordinates of the iii-th row and the jjj-th column is ijijijWalk Alone wants to play a game with

To determine who will win the game, we need to analyze the game and find a winning strategy.

First, let's consider some observations:

1. The number of chocolates placed on the board, n×mn\times mn×m, is always divisible by 2. This means that the number of turns in the game will always be even.
2. In each turn, a player can choose any row i and column j, and eat all the chocolates with coordinates (x,y) where 1≤x≤i1\leq x\leq i1≤x≤i and 1≤y≤j1\leq y\leq j1≤y≤j. This means that a player can choose to eat chocolates from any rectangular subgrid of the board, as long as it includes the top-left corner.

Based on these observations, we can devise a winning strategy for Walk Alone:

1. Walk Alone will always choose to eat chocolates that will leave an odd number of chocolates in each row and column.
2. In the first turn, Walk Alone will eat chocolates from the top-left corner of the board, leaving an odd number of chocolates in each row and column.
3. In subsequent turns, Walk Alone will mirror Kelin's move. If Kelin eats chocolates from a row i and column j, Walk Alone will eat chocolates from row n−in-i and column m−jm-j.
4. By mirroring Kelin's moves, Walk Alone ensures that an odd number of chocolates will always remain in each row and column.

Since the number of turns in the game is even, and Walk Alone's strategy guarantees that an odd number of chocolates will always remain in each row and column, Walk Alone will eat the last piece of chocolate and win the game.

Therefore, the answer is 'Walk Alone'