A. Prime Subtraction （ Educational Codeforces Round 74 (Rated for Div. 2) ）

You are given two integers xx and yy (it is guaranteed that x>yx>y). You may choose any prime integer pp and subtract it any number of times from xx. Is it possible to make xx equal to yy?

Recall that a prime number is a positive integer that has exactly two positive divisors: 11 and this integer itself. The sequence of prime numbers starts with 22, 33, 55, 77, 1111.

Your program should solve tt independent test cases.

Input

The first line contains one integer tt (1≤t≤10001≤t≤1000) — the number of test cases.

Then tt lines follow, each describing a test case. Each line contains two integers xx and yy (1≤y<x≤10181≤y<x≤1018).

Output

For each test case, print YES if it is possible to choose a prime number pp and subtract it any number of times from xx so that xx becomes equal to yy. Otherwise, print NO.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).

Example

input

Copy

4
100 98
42 32
1000000000000000000 1
41 40

output

Copy

YES
YES
YES
NO

Note

In the first test of the example you may choose p=2p=2 and subtract it once.

In the second test of the example you may choose p=5p=5 and subtract it twice. Note that you cannot choose p=7p=7, subtract it, then choose p=3p=3 and subtract it again.

In the third test of the example you may choose p=3p=3 and subtract it 333333333333333333333333333333333333 times.

特判0，1。

因为任何一个数都可以由 2 或者 2+1 组成，所以直接判断就行

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        if(n==m || n==m+1) cout<<"NO"<<endl;
        else if((n-m)%2==0 ||(n-m)%3==0||(n-m)%5==0||(n-m)%7==0 )
            cout<<"YES"<<endl;
    }
}