#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define fi first
#define se second
#define p_b push_back
#define pll pair<ll,ll>
#define pii pair<int,int>
#define m_p make_pair
#define all(x) x.begin(),x.end()
#define sset ordered_set
#define sqr(x) (x)*(x)
#define pw(x) (1ll << x)
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
const ll MAXN = 1123456;
const ll N = 2e5;
mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> void vout(T s){cout << s << endl;exit(0);}
ll ans[MAXN];
vector <ll> v[MAXN];
struct qry{
ll l, r, idx, k;
};
bool cmp(qry a, qry b){
if(a.l > b.l)return 1;
if(a.l < b.l)return 0;
return (a.idx < b.idx);
}
ll t[4 * MAXN];
void modify(ll v, ll tl, ll tr, ll pos, ll val){
if(tl == tr)t[v] = val; else{
ll tm = (tl + tr) >> 1;
if(pos <= tm)modify(v << 1, tl, tm, pos, val);
else modify((v << 1) + 1, tm + 1, tr, pos, val);
t[v] = max(t[v << 1], t[(v << 1) + 1]);
}
}
ll get(ll v, ll tl, ll tr, ll l, ll r){
if(l > r)return 0;
if(tl == l && tr == r)return t[v];
ll tm = (tl + tr) >> 1;
return max(get(v << 1, tl, tm, l, min(r, tm)), get((v << 1) + 1, tm + 1, tr, max(l, tm + 1), r));
}
int main(){
ios_base :: sync_with_stdio(0);
cin.tie(0);
ll n, m;
cin >> n >> m;
vector <ll> w(n + 1), le(n + 1);
for(int i = 1; i <= n; i++)cin >> w[i];
w[0] = 1e18;
vector <ll> st;
st.p_b(0);
for(int i = 1; i <= n; i++){
while(w[st.back()] <= w[i])st.pop_back();
le[i] = st.back();
v[le[i]].p_b(i);
st.p_b(i);
}
vector <qry> c(m);
ll uk = n;
for(int i = 0; i < m; i++){
cin >> c[i].l >> c[i].r >> c[i].k;
c[i].idx = i;
}
sort(all(c), cmp);
for(auto kek : c){
while(kek.l < uk){
for(auto j : v[uk - 1]){
modify(1, 1, n, j, w[uk - 1] + w[j]);
}
uk--;
}
ans[kek.idx] = (get(1, 1, n, kek.l, kek.r) <= kek.k);
}
for(int i = 0; i < m; i++)cout << ans[i] << "\n";
return 0;
}
