Problem: In Fire City, there are n intersections and m one-way roads. The i-th road goes from intersection a_i to b_i and has length l_i miles. 

There are q cars that may only drive along those roads. The i-th car starts at intersection v_i and has an odometer that begins at s_i, increments for each mile driven, and resets to 0 whenever it reaches t_i. Phoenix has been tasked to drive cars along some roads (possibly none) and return them to their initial intersection with the odometer showing 0.

For each car, please find if this is possible. 

A car may visit the same road or intersection an arbitrary number of times. The odometers don't stop counting the distance after resetting, so odometers may also be reset an arbitrary number of times.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the number of intersections and the number of roads, respectively.

Each of the next m lines contain three integers a_i, b_i, and l_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i; 1 ≤ l_i ≤ 10^9) — the information about the i-th road. The graph is not necessarily connected. It is guaranteed that between any two intersections, there is at most one road for each direction.

The next line contains an integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of cars.

Each of the next q lines contains three integers v_i, s_i, and t_i (1 ≤ v_i ≤ n; 0 ≤ s_i < t_i ≤ 10^9) — the initial intersection of the i-th car, the initial number on the i-th odometer, and the number at which the i-th odometer resets, respectively.

Output

Print q answers. If the i-th car's odometer may be reset to 0 by driving through some roads (possibly none) and returning to its starting intersection v_i, print YES. Otherwise, print NO.

Examples

Input


4 4
1 2 1
2 3 1
3 1 2
1 4 3
3
1 1 3
1 2 4
4 0 1


Output


YES
NO
YES


Input


4 5
1 2 1
2 3 1
3 1 2
1 4 1
4 3 2
2
1 2 4
4 3 5


Output


YES
YES

Note

The illustration for the first example is below:

<image>

In the first query, Phoenix can drive through the following cities: 1 → 2 → 3 → 1 → 2 → 3 → 1. The odometer will have reset 3 times, but it displays 0 at the end.

In the second query, we can show that there is no way to reset the odometer to 0 and return to intersection 1.

In the third query, the odometer already displays 0, so there is no need to drive through any roads.

Below is the illustration for the second example: 

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define rng(i,a,b) for(int i=int(a);i<int(b);i++)
#define rep(i,b) rng(i,0,b)
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
#define pb push_back
#define eb emplace_back
#define a first
#define b second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#define si(x) int(x.size())
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<" "<<#x<<" "<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}

template<class t> using vc=vector<t>;
template<class t> using vvc=vc<vc<t>>;

using pi=pair<int,int>;
using vi=vc<int>;

template<class t,class u>
ostream& operator<<(ostream& os,const pair<t,u>& p){
	return os<<"{"<<p.a<<","<<p.b<<"}";
}

template<class t> ostream& operator<<(ostream& os,const vc<t>& v){
	os<<"{";
	for(auto e:v)os<<e<<",";
	return os<<"}";
}

#define mp make_pair
#define mt make_tuple
#define one(x) memset(x,-1,sizeof(x))
#define zero(x) memset(x,0,sizeof(x))
#ifdef LOCAL
void dmpr(ostream&os){os<<endl;}
template<class T,class... Args>
void dmpr(ostream&os,const T&t,const Args&... args){
	os<<t<<" ";
	dmpr(os,args...);
}
#define dmp2(...) dmpr(cerr,__LINE__,##__VA_ARGS__)
#else
#define dmp2(...) void(0)
#endif

using uint=unsigned;
using ull=unsigned long long;

template<class t,size_t n>
ostream& operator<<(ostream&os,const array<t,n>&a){
	return os<<vc<t>(all(a));
}

template<int i,class T>
void print_tuple(ostream&,const T&){
}

template<int i,class T,class H,class ...Args>
void print_tuple(ostream&os,const T&t){
	if(i)os<<",";
	os<<get<i>(t);
	print_tuple<i+1,T,Args...>(os,t);
}

template<class ...Args>
ostream& operator<<(ostream&os,const tuple<Args...>&t){
	os<<"{";
	print_tuple<0,tuple<Args...>,Args...>(os,t);
	return os<<"}";
}

template<class t>
void print(t x,int suc=1){
	cout<<x;
	if(suc==1)
		cout<<"\n";
	if(suc==2)
		cout<<" ";
}

ll read(){
	ll i;
	cin>>i;
	return i;
}

vi readvi(int n,int off=0){
	vi v(n);
	rep(i,n)v[i]=read()+off;
	return v;
}

pi readpi(int off=0){
	int a,b;cin>>a>>b;
	return pi(a+off,b+off);
}

template<class t,class u>
void print(const pair<t,u>&p,int suc=1){
	print(p.a,2);
	print(p.b,suc);
}

template<class T>
void print(const vector<T>&v,int suc=1){
	rep(i,v.size())
		print(v[i],i==int(v.size())-1?suc:2);
}

string readString(){
	string s;
	cin>>s;
	return s;
}

template<class T>
T sq(const T& t){
	return t*t;
}

#define CAPITAL
void yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<"\n";
	#else
	cout<<"Yes"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void no(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<"\n";
	#else
	cout<<"No"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void possible(bool ex=true){
	#ifdef CAPITAL
	cout<<"POSSIBLE"<<"\n";
	#else
	cout<<"Possible"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}
void impossible(bool ex=true){
	#ifdef CAPITAL
	cout<<"IMPOSSIBLE"<<"\n";
	#else
	cout<<"Impossible"<<"\n";
	#endif
	if(ex)exit(0);
	#ifdef LOCAL
	cout.flush();
	#endif
}

constexpr ll ten(int n){
	return n==0?1:ten(n-1)*10;
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

int topbit(signed t){
	return t==0?-1:31-__builtin_clz(t);
}
int topbit(ll t){
	return t==0?-1:63-__builtin_clzll(t);
}
int botbit(signed a){
	return a==0?32:__builtin_ctz(a);
}
int botbit(ll a){
	return a==0?64:__builtin_ctzll(a);
}
int popcount(signed t){
	return __builtin_popcount(t);
}
int popcount(ll t){
	return __builtin_popcountll(t);
}
bool ispow2(int i){
	return i&&(i&-i)==i;
}
ll mask(int i){
	return (ll(1)<<i)-1;
}

bool inc(int a,int b,int c){
	return a<=b&&b<=c;
}

template<class t> void mkuni(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

ll rand_int(ll l, ll r) { //[l, r]
	#ifdef LOCAL
	static mt19937_64 gen;
	#else
	static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
	#endif
	return uniform_int_distribution<ll>(l, r)(gen);
}

template<class t>
void myshuffle(vc<t>&a){
	rep(i,si(a))swap(a[i],a[rand_int(0,i)]);
}

template<class t>
int lwb(const vc<t>&v,const t&a){
	return lower_bound(all(v),a)-v.bg;
}

vvc<int> readGraph(int n,int m){
	vvc<int> g(n);
	rep(i,m){
		int a,b;
		cin>>a>>b;
		//sc.read(a,b);
		a--;b--;
		g[a].pb(b);
		g[b].pb(a);
	}
	return g;
}

vvc<int> readTree(int n){
	return readGraph(n,n-1);
}

//s: the number of SCCs.
//bl[i]: SCC to which the vertex i belongs
//idx[i]: list of the vertices in the i-th SCC
//da: DAG of SCCs
//SCCs are numbered in topological order
//VERIFY: yosupo
//AOJGRL3C(except gr)
template<class t>
struct scc{
	const int n;
	const vvc<t>&g;
	vi ord,low,bl,st;
	int s,head;
	vvc<int> idx,da;
	void dfs(int v){
		ord[v]=low[v]=head++;
		st.pb(v);
		for(auto to:g[v]){
			if(ord[to]==-1){
				dfs(to);
				chmin(low[v],low[to]);
			}else if(bl[to]==-1)
				chmin(low[v],ord[to]);
		}
		if(ord[v]==low[v]){
			int c=idx.size();
			idx.eb();
			while(1){
				int a=st.back();
				st.pop_back();
				bl[a]=c;
				idx.back().pb(a);
				if(v==a)break;
			}
		}
	}
	scc(const vvc<t>&gg,bool cgr=true):n(gg.size()),g(gg),
		ord(n,-1),low(n,-1),bl(n,-1){
		head=0;
		rep(i,n)if(ord[i]==-1)
			dfs(i);
		s=idx.size();
		rep(i,n)
			bl[i]=s-1-bl[i];
		reverse(all(idx));
		//construct the graph
		if(cgr){
			vc<bool> u(s);
			da.resize(s);
			rep(i,s){
				for(auto v:idx[i]){
					for(auto to:g[v])if(bl[v]<bl[to]){
						if(!u[bl[to]])
							da[bl[v]].pb(bl[to]);
						u[bl[to]]=1;
					}
				}
				for(auto v:idx[i])
					for(auto to:g[v])
						u[bl[to]]=0;
			}
		}
	}
};
struct E{
	int to,cost;
	operator int()const{return to;}
};

int gcd(int a,int b){return b?gcd(b,a%b):a;}

const int nmax=200010;

vc<pi> g[nmax];

int dep[nmax],root[nmax],val[nmax];

void dfs(int v,int p,int d){
	if(dep[v]!=-1){
		int z=abs(dep[v]-d);
		val[root[v]]=gcd(val[root[v]],z);
		return;
	}
	dep[v]=d;
	if(p==-1)root[v]=v;
	else root[v]=root[p];
	for(auto [to,c]:g[v])
		dfs(to,v,d+c);
}

void slv(){
	int n,m;cin>>n>>m;
	using T=tuple<int,int,int>;
	vc<T> es;
	vvc<int> h(n);
	rep(_,m){
		int a,b,c;cin>>a>>b>>c;
		a--;b--;
		es.eb(a,b,c);
		h[a].pb(b);
	}
	scc<int> w(h);
	for(auto [a,b,c]:es)if(w.bl[a]==w.bl[b]){
		g[a].eb(b,c);
	}
	one(dep);
	rep(i,n)if(dep[i]==-1){
		dfs(i,-1,0);
	}
	int q;cin>>q;
	rep(_,q){
		int v,s,t;cin>>v>>s>>t;
		v--;
		int k=gcd(t,val[root[v]]);
		if((t-s)%k==0)yes(0);
		else no(0);
	}
}

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
	
	//int t;cin>>t;rep(_,t)
	slv();
}