[Click tips 23] Click here to get stones and click tips 23.

[Click tips 23] Click here to get stones and click tips 23.

Description

One day, TT was bored in the dormitory, and he played a stone game with his colleagues. However, due to limited conditions, he/they used wangzi's steamed buns As stones. The rules of the game are as follows. There is a pile of stones, the number is N (1 <= N <= 1000000), two people take several of them in turn, each time a maximum of M (1 <= M <= 1000000), the first to win the stone. We know that TT and his/her roommates are both very smart. If TT is used first, will he/she win the game?

Input

The first line is a positive integer n, indicating that n groups of test data exist.
The input contains less than 1000 groups of data. Each group contains one row of data. There are two numbers N and M, separated by spaces.

Output

Output a row for each group of data. If the first TT can Win the game, "Win" is output; otherwise, "Lose" is output (no quotation marks are required)

Analysis: this is the famous Bash Gaem)

There are only one pile of n items, and two people take things from the pile of items in turn, each time at least one, a maximum of m. The final winner wins.
If (m + 1) | n, the first hand is defeated; otherwise, the first hand is defeated.
Obviously, if n = m + 1, a maximum of m items can be taken at a time. Therefore, no matter how many items are taken by the first accessor, the latter can take the remaining items at a time, the latter wins. therefore, we send
Now we know how to win: If n = (m + 1) r + s, (r is an arbitrary natural number, s ≤ m), the first accessors should take s items, if the latter takes k (≤ m ),
Take m + 1-K, the result of the remaining (m + 1) (r-1), after maintaining such a method, then the first accessors certainly win. in a word, you must keep the multiples of (m + 1) to the opponent to win the game.

#include <stdio.h>#include <stdlib.h>int main(){int Cases;    int N,M;    scanf("%d",&Cases);    while(Cases--)    {scanf("%d %d",&N,&M);        if(N%(M+1)==0)printf("Lose\n");        elseprintf("Win\n");}return 0;}