求出lca。
再加一个cnt数组维护从根节点到i点，各种零食出现的次数。
则每种零食在任意两点u，v出现的次数是：
$res[i] = cnt[u][i] + cnt[v][i] - cnt[l][i] - cnt[fa[l][0]][i];$

#include<bits/stdc++.h>
#define endl '\n'


using namespace std;
using ll = long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int n,q;
vector<vector<int>> fa;
vector<int> dep;
vector<vector<int>> g;
vector<vector<int>> cnt;
vector<int> c;
void initlca(int x){
    for(int i=0;i<=20;i++) fa[x][i+1] = fa[fa[x][i]][i];
    for(auto y:g[x]){
        if(y==fa[x][0]) continue;
        dep[y] = dep[x]+1;
        fa[y][0] = x;
		for(int i=1;i<=20;i++){
			cnt[y][i] = cnt[x][i] + (c[y]==i);
		}

        initlca(y);
    }
}
int lca(int x,int y){
    if(dep[x]<dep[y]) swap(x,y);
    for(int d=dep[x]-dep[y],i=0;d;d>>=1,i++){
        if(d&1) x = fa[x][i];
    }
    if(x==y) return x;
    for(int i=20;i>=0;i--){
        if(fa[x][i]!=fa[y][i]) x  = fa[x][i],y = fa[y][i];
    }
    return fa[x][0];
}
void solve(){
	cin>>n>>q;
	fa = vector<vector<int>>(n+1,vector<int>(25,0));
	dep =  vector<int>(n+1);
	g = vector<vector<int>>(n+1);
	cnt = vector<vector<int>>(n+1,vector<int>(21));//从根节点到i节点的路径，包含的各种零食的数量
	c = vector<int>(n+1);
	for(int i=1;i<=n;i++) cin>>c[i];
	
	cnt[1][c[1]] = 1;
	for(int i=1;i<n;i++){
		int u,v; cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	initlca(1);

	while(q--){
		int u,v;
		cin>>u>>v;
		int l = lca(u,v);
		vector<int>	res(21,0);
		int ans = 0;
		for(int i=1;i<=20;i++){
			res[i] = cnt[u][i] + cnt[v][i] - cnt[l][i] - cnt[fa[l][0]][i];
			if(res[i] !=0) ans++;
		}
		cout<<ans<<endl;
	}

}

int main(){		
    ifstream test_file("0in.txt");
    if (test_file) {
        freopen("0in.txt", "r", stdin);
        freopen("0output.txt", "w", stdout);
    }
	std::ios::sync_with_stdio(0);std::cout.tie(0);std::cin.tie(0);
	int T = 1;
#ifdef MULTI_TEST
	cin>>T;
#endif
	while(T--){
		solve();
	}
	return 0;
}