#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2005;
const int MAXL = 5e5 + 5;
const ll INF = 1e18;
ll dp[MAXN][MAXN][2][2];
ll dp2[MAXN][2];
ll cost[MAXL][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][l] be minimum cost to collect balls i-j and then go to the end
// starting from left if k = 0, right if k = 1
// currently on left if l = 0, right if l = 1
// clearly the transitions are O(1)
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][0] = dp[i][j][k][1] = INF;
dp[0][n - 1][0][0] = dp[0][n - 1][1][1] = 0;
for (int width = n - 1; width > 0; width --) {
for (int i = 0; i < n - width; i ++) {
int j = i + width;
ll carrying = n + 1 - width;
for (int k = 0; k < 2; k ++) {
dp[i + 1][j][k][0] = min(dp[i + 1][j][k][0], dp[i][j][k][0] + (ll)abs(balls[i + 1] - balls[i]) * carrying);
dp[i + 1][j][k][1] = min(dp[i + 1][j][k][1], dp[i][j][k][0] + (ll)abs(balls[j] - balls[i]) * carrying);
dp[i][j - 1][k][0] = min(dp[i][j - 1][k][0], dp[i][j][k][1] + (ll)abs(balls[j] - balls[i]) * carrying);
dp[i][j - 1][k][1] = min(dp[i][j - 1][k][1], dp[i][j][k][1] + (ll)abs(balls[j] - balls[j - 1]) * carrying);
}
}
}
for (int i = 0; i < n; i ++)
for (int k = 0; k < 2; k ++)
dp2[i][k] = min(dp[i][i][k][0], dp[i][i][k][1]);
// now calculate best for each position
int pp = 0;
for (int i = 0; i <= l; i ++) {
cost[i][0] = cost[i][1] = INF;
while (pp < n && balls[pp] < i)
pp ++;
if (pp < n && balls[pp] == i)
cost[i][0] = dp2[pp][0], cost[i][1] = dp2[pp][1];
while (pp < n && balls[pp] == i)
pp ++;
if (pp > 0) {
cost[i][0] = min(cost[i][0], (ll)abs(balls[pp - 1] - i) * (n + 1) + dp2[pp - 1][0]);
cost[i][1] = min(cost[i][1], (ll)abs(balls[pp - 1] - i) * (n + 1) + dp2[pp - 1][1]);
}
if (pp < n) {
cost[i][0] = min(cost[i][0], (ll)abs(balls[pp] - i) * (n + 1) + dp2[pp][0]);
cost[i][1] = min(cost[i][1], (ll)abs(balls[pp] - i) * (n + 1) + dp2[pp][1]);
}
}
int q; cin >> q;
while (q --) {
int s, g, t; cin >> s >> g >> t;
t -= n;
ll best = min(cost[g][0] + abs(balls[0] - s), cost[g][1] + abs(balls[n - 1] - s));
// cout << best << endl;
if (best <= t)
cout << "Yes" << endl;
else cout << "No" << endl;
}
}
