#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll INF = 1e18;
// dp[len][i][side]: min weighted distance to cover a range of length 'len'
// starting at index 'i', ending at left (side=0) or right (side=1).
ll dp[2005][2005][2];
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int N; ll L;
cin >> N >> L;
vector<ll> X(N + 1);
for (int i = 1; i <= N; i++) cin >> X[i];
sort(X.begin() + 1, X.end());
int Q; cin >> Q;
while (Q--) {
ll S, G, T;
cin >> S >> G >> T;
// Reset DP for each scenario (only feasible for Q <= 10)
// If Q is large, we need the O(N) approach below.
for (int i = 1; i <= N; i++) {
dp[1][i][0] = dp[1][i][1] = abs(S - X[i]);
}
for (int len = 1; len < N; len++) {
for (int i = 1; i + len - 1 <= N; i++) {
int j = i + len - 1;
if (dp[len][i][0] >= INF && dp[len][i][1] >= INF) continue;
// Move from left end (i) to i-1 or j+1
if (i > 1) {
dp[len+1][i-1][0] = min(dp[len+1][i-1][0], dp[len][i][0] + (ll)(len + 1) * (X[i] - X[i-1]));
}
if (j < N) {
dp[len+1][i][1] = min(dp[len+1][i][1], dp[len][i][0] + (ll)(len + 1) * (X[j+1] - X[i]));
}
// Move from right end (j) to i-1 or j+1
if (i > 1) {
dp[len+1][i-1][0] = min(dp[len+1][i-1][0], dp[len][i][1] + (ll)(len + 1) * (X[j] - X[i-1]));
}
if (j < N) {
dp[len+1][i][1] = min(dp[len+1][i][1], dp[len][i][1] + (ll)(len + 1) * (X[j+1] - X[j]));
}
}
}
ll min_travel = min(dp[N][1][0] + (ll)(N + 1) * abs(X[1] - G),
dp[N][1][1] + (ll)(N + 1) * abs(X[N] - G));
if (N + min_travel <= T) cout << "Yes\n";
else cout << "No\n";
}
return 0;
}
