纯模板，思路也很简单，用SG函数推一下就能发现。对于当前的状态x，他的SG值就是mex{SG[x-m],……,SG[x-1]}。

然后手动推一下打表就能发现，当n%(m+1)==0的时候，无论先手选择什么，后手必定有一种选择能够让两个人选择的石头数等于m+1,所以后手必胜

反之，先手就可以拿走n%(m+1)个石头，这样子自己成为上一个状态的后手，所以必胜

代码如下:

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//         g ....,i, OS   ,X..7HrL   :        1r..:s. L:,;....,:. K2:.L
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// , , :QBB1 ;:7v:vJ:    Jv.i1EgBgJi,     .,::.gDZKPHGBs,;  .BQMDU5GQBBB7
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// , .:. . DQK.r.7; r.   :gBDZpZPEpZ6gMa6RMBBQgEKZ6DDMBr,..::BEZpEGDZPPG
//   si,: :2QBac;..,r     PQgPPPEZOZDGDRRDDEGpZpOOgORBH,: ..:Qg6ZZp2JsO;
//  Q: :J7 7      .,.     6RQQDD6ZpZpZ6Z6ZpZ6gOgDOGRDR  :.  .MMOpX52UKDr
// BB7 1  .:UMOPXr        gQgEgDDZZpE6E6Z6ZPZGOpDRDUSOEHBBL  7BRP5HXXXDJ.
// GXB1  .: UBMQBBg       JB6pKpPG6O6G6EZEpEPp6MGSwPEDRQgMBP  :RPXUX6PQa;
// gXOR .,,pgGE6ODBB,      QQOZ6RMBBBQMgDZZ6ZGE21HOEZpEpEZgB,  :ZZppBQs,v
// gwXBr,;EBEEpGZGGBQ:     .RMgK7r:::i7s1HPZHDHLEGPpKpK6PpZBs .D;g6GMZ  ;

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ll long long
#define lowbit(x) x&(-x)

const int MOD=1e9 + 7;
const int N=1e6+10;

int qpow(int num,int k) {
    int res=1;
    while(k) {
        if(k%2) res = (res * num) % MOD;
        num = (num * num) % MOD;
        k/=2 ;
    }
    return res % MOD;
}

int inv(int x) {
    return qpow(x,MOD-2);
}
 
int lcm(int x,int y) {
    return x * y / __gcd(x,y);
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int Base = uniform_int_distribution<>(8e8,9e8)(rng); 

int a[N];
int b[N];

void solve() {
    int n,m;
    cin>>n>>m;
    if(n % (m + 1)) cout<<"YES\n";
    else cout<<"NO\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t=1;
    cin>>t;

    while(t--) {
        solve();
    }
    return 0;
}