// Om Namah Shivaya
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a, b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a, b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
guy j has to cross guy i (i < j) if:
a[j] < a[i]
for all guys that must cross each other, find if a[i]+a[j] <= k
rewriting the problem:
for given (l,r), find the max val of a[i] + a[j], s.t:
i < j and a[i] > a[j]
i.e find max sum inversion pair in range
we can try to use sweepline
sweep over r, maintaining the max ans for all l
so when answering a query, we can just find the max sum over all l in range [l,r] and check if it's > k
what changes when we move from r-1 to r
a[r] forms an inversion pair only with a[i] > a[r], i < r
let the indices of all guys < r with val > a[r] in descending order of indices be:
i1 > i2 > ... > ik
(i1,r) forms an inversion pair, so we can upd the val at i1 to a[i1] + a[r]
(i2,r) also forms an inversion pair
there are 2 cases to consider for i2:
1) a[i2] <= a[i1]: (i1,r) gives better result than (i2,r), so no need to update
2) a[i2] > a[i1]: (i2,i1) gives better result than (i2,r), so no need to update
similar arguments apply for i3,i4,...,ik
so we just have to upd the val at i1
i1 = next greater value to the left of a[r]
which can be found using a stack
updates and range max queries can be handled with a segtree
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
template<typename T>
struct segtree {
// https://codeforces.com/blog/entry/18051
/*=======================================================*/
struct data {
ll a;
};
data neutral = {-inf2};
data merge(data &left, data &right) {
data curr;
curr.a = max(left.a, right.a);
return curr;
}
void create(int i, T v) {
}
void modify(int i, T v) {
amax(tr[i].a, v);
}
/*=======================================================*/
int n;
vector<data> tr;
segtree() {
}
segtree(int siz) {
init(siz);
}
void init(int siz) {
n = siz;
tr.assign(2 * n, neutral);
}
void build(vector<T> &a, int siz) {
rep(i, siz) create(i + n, a[i]);
rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
}
void pupd(int i, T v) {
modify(i + n, v);
for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
}
data query(int l, int r) {
data resl = neutral, resr = neutral;
for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
if (l & 1) resl = merge(resl, tr[l++]);
if (!(r & 1)) resr = merge(tr[r--], resr);
}
return merge(resl, resr);
}
};
void solve(int test_case)
{
ll n,m; cin >> n >> m;
vector<ll> a(n+5);
rep1(i,n) cin >> a[i];
stack<ll> st;
vector<ll> nge(n+5,-1);
rev(i,n,1){
while(!st.empty() and a[i] > a[st.top()]){
nge[st.top()] = i;
st.pop();
}
st.push(i);
}
vector<array<ll,3>> queries[n+5];
rep1(i,m){
ll l,r,k; cin >> l >> r >> k;
queries[r].pb({l,k,i});
}
segtree<ll> seg(n+5);
vector<ll> ans(m+5);
rep1(r,n){
ll j = nge[r];
if(j != -1){
seg.pupd(j, a[r] + a[j]);
}
for(auto [l,k,id] : queries[r]){
ll mx = seg.query(l,r).a;
if(mx > k) ans[id] = 0;
else ans[id] = 1;
}
}
rep1(i,m) cout << ans[i] << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
