Problem: In Homer's country, there are n cities numbered 1 to n and they form a tree. That is, there are (n-1) undirected roads between these n cities and every two cities can reach each other through these roads. 

Homer's country is an industrial country, and each of the n cities in it contains some mineral resource. The mineral resource of city i is labeled a_i. 

Homer is given the plans of the country in the following q years. The plan of the i-th year is described by four parameters u_i, v_i, l_i and r_i, and he is asked to find any mineral resource c_i such that the following two conditions hold: 

  * mineral resource c_i appears an odd number of times between city u_i and city v_i; and 
  * l_i ≤ c_i ≤ r_i. 



As the best friend of Homer, he asks you for help. For every plan, find any such mineral resource c_i, or tell him that there doesn't exist one.

Input

The first line contains two integers n (1 ≤ n ≤ 3 ⋅ 10^5) and q (1 ≤ q ≤ 3 ⋅ 10^5), indicating the number of cities and the number of plans.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

Then the i-th line of the following (n-1) lines contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n) with x_i ≠ y_i, indicating that there is a bidirectional road between city x_i and city y_i. It is guaranteed that the given roads form a tree.

Then the i-th line of the following q lines contains four integers u_i, v_i, l_i, r_i (1 ≤ u_i ≤ n, 1 ≤ v_i ≤ n, 1 ≤ l_i ≤ r_i ≤ n), indicating the plan of the i-th year.

Output

Print q lines, the i-th of which contains an integer c_i such that 

  * c_i = {-1} if there is no such mineral resource that meets the required condition; or 
  * c_i is the label of the chosen mineral resource of the i-th year. The chosen mineral resource c_i should meet those conditions in the i-th year described above in the problem statement. If there are multiple choices of c_i, you can print any of them. 

Example

Input


6 8
3 2 1 3 1 3
1 2
1 3
2 4
2 5
4 6
3 5 1 1
3 5 1 3
3 5 1 3
1 1 2 2
1 1 3 3
1 4 1 5
1 6 1 3
1 6 1 3


Output


-1
2
3
-1
3
2
2
3

Note

In the first three queries, there are four cities between city 3 and city 5, which are city 1, city 2, city 3 and city 5. The mineral resources appear in them are mineral resources 1 (appears in city 3 and city 5), 2 (appears in city 2) and 3 (appears in city 1). It is noted that 

  * The first query is only to check whether mineral source 1 appears an odd number of times between city 3 and city 5. The answer is no, because mineral source 1 appears twice (an even number of times) between city 3 and city 5. 
  * The second and the third queries are the same but they can choose different mineral resources. Both mineral resources 2 and 3 are available. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using db = long double; // or double, if TL is tight
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;

#define tcT template<class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back 
#define pf push_front
#define rtn return

#define lb lower_bound
#define ub upper_bound 
tcT> int lwb(V<T>& a, const T& b) { return int(lb(all(a),b)-bg(a)); }

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define rep(a) F0R(_,a)
#define each(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const db PI = acos((db)-1);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;

// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { // assert(x >= 0); // make C++11 compatible until USACO updates ...
	return x == 0 ? 0 : 31-__builtin_clz(x); } // floor(log2(x)) 
constexpr int p2(int x) { return 1<<x; }
constexpr int msk2(int x) { return p2(x)-1; }

ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(it); } // element that doesn't exist from (multi)set

// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(V<T>& x) { each(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { each(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	// #ifdef LOCAL
	// 	return b ? "true" : "false"; 
	// #else 
	return ts((int)b);
	// #endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(V<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;
	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void setIO(str s = "") {
	unsyncIO(); setPrec();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) setIn(s+".in"), setOut(s+".out"); // for USACO
}

#define ints(...) int __VA_ARGS__; re(__VA_ARGS__);

class Fenwick{
public:
	vector<int> li;
	int nn;
	Fenwick(int x){
		nn = x+5;
		li.resize(nn);
	}
	void add(int ind, int val){
		ind+=2;
		while(ind<nn){
			li[ind] ^= val;
			ind += ind & (-ind);
		}
	}
	int sum(int ind){
		ind += 2;
		int ret = 0LL;
		while(ind>0){
			ret ^= li[ind];
			ind -= ind & (-ind);
		}
		return ret;
	}
	void res(){
		for(int i = 0; i<li.size(); i++){
			li[i] = 0LL;
		}
	}

};

struct LCA {
public:
	int N; vector<vi> par, adj_lca; vi depth;
	void init(int _N) {  N = _N;
		int d = 1; while ((1<<d) < N) d ++;
		par.assign(d,vi(N)); adj_lca.rsz(N); depth.rsz(N);
	}
	void ae(int x, int y) { adj_lca[x].pb(y), adj_lca[y].pb(x); }
	void gen(int R = 0) { par[0][R] = R; dfs(R); }
	void dfs(int x = 0) {
		FOR(i,1,sz(par)) par[i][x] = par[i-1][par[i-1][x]];
		each(y,adj_lca[x]) if (y != par[0][x]) 
			depth[y] = depth[par[0][y]=x]+1, dfs(y);
	}
	int jmp(int x, int d) {
		F0R(i,sz(par)) if ((d>>i)&1) x = par[i][x];
		return x; }
	int lca(int x, int y) {
		if (depth[x] < depth[y]) swap(x,y);
		x = jmp(x,depth[x]-depth[y]); if (x == y) return x;
		R0F(i,sz(par)) {
			int X = par[i][x], Y = par[i][y];
			if (X != Y) x = X, y = Y;
		}
		return par[0][x];
	}
	int dist(int x, int y) { // # edges on path
		return depth[x]+depth[y]-2*depth[lca(x,y)]; }
};

vector<int> times[600001];
int pre[300001];
int results[300001];
int og_results[300001];
vector<pi> queries[600002];
int lo[300001];
int hi[300001];
bool has[300001];
int n, q;
int u[300001];
int v[300001];
int l[300001];
int r[300001];
int a[300001];
int lca_val[300001];

int hsh_val[300001];
vector<int> adj[300001];
int point = 1;
void go(int now, int from){
	times[a[now]].pb(point++);
	pre[now] = point-1;
	for(int i = 0; i<adj[now].size(); i++){
		int to = adj[now][i];
		if(to==from){
			continue;
		}
		go(to,now);
	}
	times[a[now]].pb(point++);
}
Fenwick fen = Fenwick(600002);
LCA lca_ds;
void reset_queries(){
	fen.res();
	for(int i = 0; i<point; i++){
		queries[i].clear();
	}
	for(int i = 0; i<=q; i++){
		results[i] = 0LL;
	}
}
void make_query_start(int st, int ind){
	// do after
	queries[st-1].pb(mp(pre[u[ind]],ind));
	queries[st-1].pb(mp(pre[v[ind]],ind));

}
void make_query_end(int st, int en, int ind){
	// do after
	queries[en].pb(mp(pre[u[ind]],ind));
	queries[en].pb(mp(pre[v[ind]],ind));
	int got_lca = lca_val[ind];
	if(a[got_lca]>=st && a[got_lca]<=en){
		results[ind] ^= hsh_val[a[got_lca]];
	}

}
void make_query(int st, int en, int ind){
	// do after
	queries[en].pb(mp(pre[u[ind]],ind));
	queries[en].pb(mp(pre[v[ind]],ind));
	int got_lca = lca_val[ind];
	if(a[got_lca]>=st && a[got_lca]<=en){
		results[ind] ^= hsh_val[a[got_lca]];
	}
	queries[st-1].pb(mp(pre[u[ind]],ind));
	queries[st-1].pb(mp(pre[v[ind]],ind));

}
void handle_queries(){
	for(int i = 0; i<point; i++){
		for(int j = 0; j<times[i].size(); j++){
			fen.add(times[i][j],hsh_val[i]);
		}
		for(int j = 0; j<queries[i].size(); j++){
			results[queries[i][j].s] ^= fen.sum(queries[i][j].f);
		}
	}
}
int main() {
	setIO();
	cin >> n >> q;
	for(int i = 0; i<n; i++){
		cin >> a[i];
	}
	lca_ds.init(n);
	for(int i = 1; i<n; i++){
		int a, b;
		cin >> a >> b;
		a--;
		b--;
		adj[a].pb(b);
		adj[b].pb(a);
		lca_ds.ae(a,b);
	}
	lca_ds.gen();
	go(0,-1);
	// 1000006019
	// 1000004023
	// 900004043
	// 910004063
	// 910004087
	// 910004099
	for(int i = 1; i<=n; i++){
		ll a = (ll)i*900004043LL+910004063LL;
		a %= 1000006019LL;
		hsh_val[i] = (int)a;
	}
	reset_queries();
	for(int i = 0; i<q; i++){
		cin >> u[i] >> v[i] >> l[i] >> r[i];
		u[i]--;
		v[i]--;
		lca_val[i] = lca_ds.lca(u[i],v[i]);
		make_query(l[i],r[i],i);
	}
	handle_queries();
	for(int i = 0; i<q; i++){
		if(results[i]!=0LL){
			has[i] = true;
			lo[i] = l[i];
			hi[i] = r[i];
		}
		else{
			has[i] = false;
		}
	}
	reset_queries();
	for(int i = 0; i<q; i++){
		if(has[i]){
			make_query_start(l[i],i);
		}
	}
	handle_queries();
	for(int i = 0; i<q; i++){
		if(has[i]){
			og_results[i] = results[i];
		}
	}
	while(true){
		int cnt = 0;
		reset_queries();
		for(int i = 0; i<q; i++){
			if(has[i] && lo[i]!=hi[i]){
				make_query_end(l[i],(lo[i]+hi[i])/2,i);
				// make_query(lo[i],(lo[i]+hi[i])/2,i);
				cnt++;
			}
		}
		if(cnt==0){
			break;
		}
		handle_queries();
		for(int i = 0; i<q; i++){
			if(has[i] && lo[i]!=hi[i]){
				// if(results[i] != 0){
				if(results[i] != og_results[i]){
					hi[i] = (lo[i]+hi[i])/2;
				}
				else{
					lo[i] = (lo[i]+hi[i])/2+1;
				}
			}
		}
	}
	for(int i = 0; i<q; i++){
		if(!has[i]){
			cout << -1 << "\n";
		}
		else{
			cout << lo[i] << "\n";
		}
	}
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/