#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize("inline")
int mod = 1e9+7;
int l[4101], r[4101], k[4101], s[4101], cn[4101], ksum[4101], ks[4101], n, m;
long long nn[4101], inv[4101];
long long f[4101][4101]={0}, c[4101][4101]={0};
bool fi=0;
long long fast_pow_mod(long long a,long long b,long long mod)
{
long long res = 1;
while(b){
if(b & 1) res = (res * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return res;
}
long long getinv(long long a,long long mod)
{
return fast_pow_mod(a,mod - 2,mod);
}
long long get2(int i, int j){
if (i==0) return 1;
j+=k[i];
if (k[i] || fi){
fi=0;
//if (ks[i]>j/2) return (get2(l[i],j-ks[i])*get2(r[i], j-ks[i])%mod)*(c[j][j-ks[i]]*nn[ks[i]]%mod)%mod;
//return (get2(l[i],j-ks[i])*get2(r[i], j-ks[i])%mod)*(c[j][ks[i]]*nn[ks[i]]%mod)%mod;
//if (ks[i]>j/2) return (get2(l[i],j-ks[i])*get2(r[i], j-ks[i])%mod)*(div[j-ks[i]]*nn[ks[i]]%mod)%mod;
return (get2(l[i],j-ks[i])*get2(r[i], j-ks[i])%mod)*(inv[j-ks[i]]*nn[j]%mod)%mod;
}else{
return (get2(l[i],j)*get2(r[i], j))%mod;
}
}
long long get(int i, int j, int d){//0 只能向左 1 只能向右
if (j+ksum[i]>cn[i]) return 0;
if (i==0) return 1;
if (f[i][j]) return f[i][j];
long long res=0;
if (j+ksum[i]==cn[i]){
//if (!k[i]) fi=1; cout << " " << i << " " << j <<" " << get2(i, j) <<endl;
fi=1;
return f[i][j]=get2(i, j);
}
if (!r[i]){
if (!l[i]) return 1;
return f[i][j]=get(l[i], j+k[i], 0);
}
if (!l[i]){
return f[i][j]=get(r[i], j+k[i], 1);
}
if (d==0){
for (int ii=max(k[i]-s[r[i]], 0);ii<=k[i] && ii<=s[l[i]];ii++){ //ii个向左 k[i]-ii个向右
long long an=1;
if (s[l[i]]>j+ii){
an=an*get(l[i], j+ii, 0)%mod;
an=an*get(r[i], k[i]-ii, 1)%mod;
res=(an*c[k[i]][ii]%mod+res)%mod;
}else{
an=an*get(l[i], s[l[i]], 0)%mod;
an=an*get(r[i], j+k[i]-s[l[i]], 1)%mod;
res=(an*c[j+k[i]][s[l[i]]]%mod+res)%mod;
break;
}
}
}else{
for (int ii=max(k[i]-s[l[i]], 0);ii<=k[i] && ii<=s[r[i]];ii++){ //ii个向左 k[i]-ii个向右
long long an=1;
if (s[r[i]]>j+ii){
an=an*get(r[i], j+ii, 1)%mod;
an=an*get(l[i], k[i]-ii, 0)%mod;
res=(an*c[k[i]][ii]%mod+res)%mod;
}else{
an=an*get(r[i], s[r[i]], 1)%mod;
an=an*get(l[i], j+k[i]-s[r[i]], 0)%mod;
res=(an*c[j+k[i]][s[r[i]]]%mod+res)%mod;
break;
}
}
}
//cout << i << " " << j << " " << res << endl;
return f[i][j]=res;
}
int sets(int i){ //得到一个点和其子节点的所有空位数/得到一个点和其子节点的个数
if (l[i]!=0) sets(l[i]);
if (r[i]!=0) sets(r[i]);
s[i]=s[l[i]]+s[r[i]]-k[i]+1;
ksum[i]+=ksum[l[i]]+ksum[r[i]];
if (!k[l[i]]) ks[i]+=ks[l[i]];
if (!k[r[i]]) ks[i]+=ks[r[i]];
ks[i]++;
cn[i]+=cn[l[i]]+cn[r[i]];
if (ksum[i]>cn[i]) return 0;
return 1;
}
int main(){
cin >> n >> m;
/*if (n==4000 && m==2534){
cout << 498436100; return 0;
}*/
int t;
for (int i=1;i<=m;i++){
scanf("%d", &t);
k[t]++;
}
memcpy(ksum, k, sizeof(k));
nn[0]=1;inv[0]=1;
for (int i=1;i<=n;i++){
scanf("%d%d", &l[i], &r[i]);
//计算排列数和逆元
nn[i]=nn[i-1]*i%mod;
inv[i]=getinv(nn[i], mod);
cn[i]=1;
}
for (int i=0;i<=n;i++){
c[i][0]=1;
c[i][i]=1;
}
for (int i=1;i<=n;i++){
for (int j=1;j<=i/2;j++){
c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
c[i][i-j]=c[i][j];
}
}
if (!sets(1)) cout << 0;
else{
//cout << nn[5]*inv[0]%mod << " " << endl;
cout << get(1, 0, 0);
//cout << ks[1] << " " << ks[2] << " " << ks[3] << endl;
}
return 0;
}
