#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2005;
const ll INF = 1e18;
ll dp[MAXN][MAXN][2];
int main() {
int n; cin >> n;
int l; cin >> l;
vector<int> balls(n);
for (auto &a : balls) cin >> a;
sort(balls.begin(), balls.end());
// let dp[i][j][k] be minimum cost to collect balls i-j and then go to the end
// starting from left if k = 0, right if k = 1
// clearly the transitions are O(1)
int q; cin >> q;
while (q --) {
int s, g, t; cin >> s >> g >> t;
t -= n;
for (int i = 0; i < n; i ++)
for (int j = 0; j < n; j ++)
for (int k = 0; k < 2; k ++)
dp[i][j][k] = INF;
for (int i = 0; i < n; i ++) {
dp[i][i][0] = dp[i][i][1] = (ll)abs(balls[i] - g) * (ll)(n + 1);
}
for (int width = 1; width < n; width ++) {
for (int i = 0; i < n - width; i ++) {
int j = i + width;
ll carrying = n + 1 - width;
dp[i][j][0] = min(dp[i][j][0], dp[i + 1][j][0] + (ll)abs(balls[i + 1] - balls[i]) * carrying);
dp[i][j][0] = min(dp[i][j][0], dp[i + 1][j][1] + (ll)abs(balls[j] - balls[i]) * carrying);
dp[i][j][1] = min(dp[i][j][1], dp[i][j - 1][0] + (ll)abs(balls[j] - balls[i]) * carrying);
dp[i][j][1] = min(dp[i][j][1], dp[i][j - 1][1] + (ll)abs(balls[j] - balls[j - 1]) * carrying);
}
}
ll best = min(dp[0][n - 1][0] + abs(balls[0] - s), dp[0][n - 1][1] + abs(balls[n - 1] - s));
// cout << best << " " << dp[0][n - 2][1] << endl;
if (best <= t)
cout << "Yes" << endl;
else cout << "No" << endl;
}
}
