#include <bits/stdc++.h>
using namespace std;
class SegTree {
public:
int sz;
vector<int> tree;
SegTree(int sz) : sz(sz), tree(2 * sz, 0) {}
void Modify(int p, int x) {
for (p += sz; p > 0; p /= 2) tree[p] = max(tree[p], x);
}
int Query(int l, int r) {
int res = 0;
for (l += sz, r += sz + 1; l < r; l /= 2, r /= 2) {
if (l & 1) res = max(res, tree[l++]);
if (r & 1) res = max(res, tree[--r]);
}
return res;
}
};
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int N, Q;
cin >> N >> Q;
vector<int> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
vector<int> ans(Q);
vector<vector<array<int, 3>>> query(N);
for (int i = 0; i < Q; i++) {
int l, r, k;
cin >> l >> r >> k;
l--, r--;
query[l].push_back({r, k, i});
}
// Solution:
//
// For a range [L, R], we want to pick
// i, j, s.t. L <= i < k <= R, A[i] > A[j],
// where A[i] + A[j] is maximal. Then, if
// A[i] + A[j] > k, the books cannot be sorted;
// otherwise, it can.
//
// Note that for i < j, A[i] > A[j], since we can
// only swap adjacent books, at some point in sorting
// A[i] and A[j] must be adjacent, and we must swap them.
//
// Let's sweep from right to left. Note that for i < j,
// A[i] > A[j]:
// - If there exist i < k < j, s.t. A[j] < A[i] < A[k], the pair
// (k, j) is strictly better.
// - If there exist i < k < j, s.t. A[i] > A[k] > A[j], the pair
// (i, k) is strictly better.
//
// Note that this means, for all (i, j), it is optimal if all indices
// i < k < j, either A[k] < min(A[i], A[j]) must hold.
//
// This means, we only need to care about the nearest element to the left
// of j which is greater than A[j]. This is only O(N), and can be computed
// easily.
//
// We can answer everything offline in O((N + Q) log N).
vector<vector<int>> candidates(N);
{ // Compute candidates, (i, j) is good if i is nearest element greater than A[j] to the right
vector<int> st;
for (int i = 0; i < N; i++) {
while (!st.empty() && A[st.back()] <= A[i]) {
st.pop_back();
}
if (!st.empty()) {
candidates[st.back()].emplace_back(i);
}
st.emplace_back(i);
}
}
SegTree seg(N);
for (int l = N - 1; l >= 0; l--) {
for (auto r : candidates[l]) {
seg.Modify(r, A[l] + A[r]);
}
for (auto [r, k, q] : query[l]) {
ans[q] = seg.Query(l, r) <= k;
}
}
for (int i = 0; i < Q; i++) {
cout << ans[i] << '\n';
}
return 0;
}
