$k$的范围非常小, $O(n2^k)$求出状态最多为$S$的路径数, 然后容斥.
#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif
int n, k, mx, sum;
int a[N], dp[N], vis[N];
vector<int> g[N];
void dfs(int x, int fa, int S) {
if ((a[x]|S)!=S||vis[x]) return;
++sum, vis[x]=1;
for (int y:g[x]) if (y!=fa) dfs(y,x,S);
}
int main() {
scanf("%d%d", &n, &k);
REP(i,1,n) a[i]=1<<rd()-1;
REP(i,2,n) {
int u=rd(),v=rd();
g[u].pb(v),g[v].pb(u);
}
int mx = (1<<k)-1;
REP(S,1,mx) {
REP(i,1,n) vis[i]=0;
REP(i,1,n) if (!vis[i]) {
sum = 0, dfs(i,0,S);
dp[S] = ((ll)sum*(sum-1)%P*(P+1)/2+sum+dp[S])%P;
}
}
int ans = 0;
REP(S,1,mx) {
int cnt = 0;
for (int j=(S-1)&S; j; j=(j-1)&S) dp[S]-=dp[j];
ans = ((ll)ans+qpow(131,__builtin_popcount(S))*dp[S])%P;
}
if (ans<0) ans+=P;
printf("%d\n", ans);
}