//#pragma GCC optimize("Ofast,unroll-loops,O3")
//#pragma GCC target("avx,avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,fma,tune=native")
#include <bits/stdc++.h>
#define int long long
#define pii pair<int, int>
#define x first
#define y second
#define N 2015
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, L, arr[N], dpl[N][N][2], dpr[N][N][2];
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cin >> n >> L;
for (int i = 1; i <= n; i++)
cin >> arr[i];
sort(arr + 1, arr + 1 + n);
for (int i = 0; i <= n + 1; i++)
for (int j = 0; j <= n + 1; j++){
dpl[i][j][0] = dpl[i][j][1] = 5e18;
dpr[i][j][0] = dpr[i][j][1] = 5e18;
}
dpl[1][n][0] = 0, dpl[1][n][1] = arr[n] - arr[1];
dpr[1][n][0] = arr[n] - arr[1], dpr[1][n][1] = 0;
for (int j = n - 1; j >= 1; j--){
for (int i = 1; i <= n - j + 1; i++){
dpl[i][j][0] = dpl[i - 1][j + 1][0] + (arr[i] - arr[i - 1]) * (n - j + 1);
dpl[i][j][1] = dpl[i][j + 1][1] + (arr[i + j] - arr[i + j - 1]) * (n - j + 1);
dpl[i][j][0] = min(dpl[i][j][0], dpl[i][j][1] + (arr[i + j - 1] - arr[i]) * (n - j + 1));
dpl[i][j][1] = min(dpl[i][j][1], dpl[i][j][0] + (arr[i + j - 1] - arr[i]) * (n - j + 1));
dpr[i][j][0] = dpr[i - 1][j + 1][0] + (arr[i] - arr[i - 1]) * (n - j + 1);
dpr[i][j][1] = dpr[i][j + 1][1] + (arr[i + j] - arr[i + j - 1]) * (n - j + 1);
dpr[i][j][0] = min(dpr[i][j][0], dpr[i][j][1] + (arr[i + j - 1] - arr[i]) * (n - j + 1));
dpr[i][j][1] = min(dpr[i][j][1], dpr[i][j][0] + (arr[i + j - 1] - arr[i]) * (n - j + 1));
}
}
// cerr << "from left :\n";
// for (int i = 1; i <= n; i++){
// for (int j = i; j <= n; j++){
// cerr << i << " ~ " << j << " -> " << dpl[i][j - i + 1][0] << ", " << dpl[i][j - i + 1][1] << '\n';
// }
// }
// cerr << "from right :\n";
// for (int i = 1; i <= n; i++){
// for (int j = i; j <= n; j++){
// cerr << i << " ~ " << j << " -> " << dpr[i][j - i + 1][0] << ", " << dpr[i][j - i + 1][1] << '\n';
// }
// }
int q; cin >> q;
while (q--){
int s, t, T, ans = 5e18;
cin >> s >> t >> T;
for (int i = 1; i <= n; i++){
int ans1 = dpl[i][1][0] + abs(arr[i] - t) * (n + 1) + abs(arr[1] - s);
int ans2 = dpr[i][1][0] + abs(arr[i] - t) * (n + 1) + abs(arr[n] - s);
ans = min({ans, ans1, ans2});
}
if (ans + n <= T)
cout << "Yes\n";
else cout << "No\n";
// cout << ans + n << " " << T << '\n';
}
}
