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ZB loves watching RunningMan! There's a game in RunningMan called 100 vs 100.

There are two teams, each of many people. There are 3 rounds of fighting, in each round the two teams send some people to fight. In each round, whichever team sends more people wins, and if the two teams send the same amount of people, RunningMan team wins. Each person can be sent out to only one round. The team wins 2 rounds win the whole game. Note, the arrangement of the fighter in three rounds must be decided before the whole game starts.

We know that there are N people on the RunningMan team, and that there are M people on the opposite team. Now zb wants to know whether there exists an arrangement of people for the RunningMan team so that they can always win, no matter how the opposite team arrange their people.

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, there's one line consists of two integers N and M. (1 <= N, M <= 10^9).

Output

For each test case, Output "Yes" if there exists an arrangement of people so that the RunningMan team can always win. "No" if there isn't such an arrangement. (Without the quotation marks.)

Sample Input

2100 100200 100

Sample Output

NoYes

Hint

In the second example, the RunningMan team can arrange 60, 60, 80 people for the three rounds. No matter how the opposite team arrange their 100 people, they cannot win.

题目大意：两个队进行3场战斗，每场两个队都配一些人，每场比赛中，如果两个队人数相等，Running man赢，每个人只能参加一次，三局两赢制，不论Runningman怎么安排人，他们都能赢，则输出Yes，否则输出No。

#include<iostream>#include<cstdio>using namespace std;int main(){    int T;    scanf("%d", &T);    while(T--)    {        int n,m,a,b,c,t;        scanf("%d%d",&n,&m);            a=n/3;            b=a;            t=n%3;            if(t==2)                b++;            if(b+a+2>m)            printf("Yes\n");            else            printf("No\n");    }    return 0;}