Problem:
Score : 600 points

Problem Statement

Ringo Mart, a convenience store, sells apple juice.

On the opening day of Ringo Mart, there were A cans of juice in stock in the morning.
Snuke buys B cans of juice here every day in the daytime.
Then, the manager checks the number of cans of juice remaining in stock every night.
If there are C or less cans, D new cans will be added to the stock by the next morning.

Determine if Snuke can buy juice indefinitely, that is, there is always B or more cans of juice in stock when he attempts to buy them.
Nobody besides Snuke buy juice at this store.

Note that each test case in this problem consists of T queries.

Constraints

1 \leq T \leq 300

1 \leq A, B, C, D \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

T
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_T B_T C_T D_T

In the i-th query, A = A_i, B = B_i, C = C_i, D = D_i.

Output

Print T lines. The i-th line should contain Yes if Snuke can buy apple juice indefinitely in the i-th query; No otherwise.

Sample Input 1

14
9 7 5 9
9 7 6 9
14 10 7 12
14 10 8 12
14 10 9 12
14 10 7 11
14 10 8 11
14 10 9 11
9 10 5 10
10 10 5 10
11 10 5 10
16 10 5 10
1000000000000000000 17 14 999999999999999985
1000000000000000000 17 15 999999999999999985

Sample Output 1

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes

In the first query, the number of cans of juice in stock changes as follows: (D represents daytime and N represents night.)

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 6
→D x

In the second query, the number of cans of juice in stock changes as follows:

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 15
→D 8
→N 8
→D 1
→N 10
→D 3
→N 12
→D 5
→N 14
→D 7
→N 7
→D 0
→N 9
→D 2
→N 11
→D …

and so on, thus Snuke can buy juice indefinitely.

Sample Input 2

24
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1

Sample Output 2

No
No
No
No
No
No
Yes
Yes
No
No
No
No
Yes
Yes
Yes
No
No
No
Yes
Yes
Yes
No
No
No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
#define thekingisborn ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ld  long double
#define f first
#define se second
#define pb push_back
#define maxn 200005
#define mod 1000000007
#define pi 3.14159265359

using namespace std;

typedef pair<int,int>pii;
typedef pair<ll,ll>pll;

/*ll power(ll a,ll n)
{
    if(n == 1)return a%mod;
    if(n == 0)return 1;
    a = a%mod;
    if(n&1)return ((a%mod)*(power((a*a)%mod,n>>1))%mod)%mod;
    return power((a*a)%mod,n>>1)%mod;
}
ll gcd(ll a,ll b)
{
    if(a%b == 0)return b;
    return gcd(b,a%b);
}*/
/*int root(int x)
{
    while(par[x] != x)
    x = par[x];
    return x;
}
void dsu(int x,int y)
{
  if(root(x) == root(y))return;
  par[root(y)] = root(x);
  return;
}
void dfs(int x)
{
    done[x] = true;
    for(int i=0;i<int(adj[x].size());++i)
    {
        int y = adj[x][i];
        if(!done1[y])
        dfs(y);
    }
}*/
int t;
ll a,b,c,d;

int main()
{
    thekingisborn;
    cin >>t;
    for(int i=1;i<=t;++i)
    {
        cin >> a >> b>> c >> d;
        ll cur = a,j=0;
        bool f=0;
        while(j<100000)
        {
            if(cur < b){cout << "No\n";f=1;break;}
            cur = cur%b;
            if(cur <= c){cur += d;}
            ++j;
        }
        if(!f)cout << "Yes\n";
    }

    return 0;
}