#include <bits/stdc++.h>
using namespace std;
#define int long long
vector<int> adj[100005];
int dp[100005][2][2]; //end with this state instead, whether you are choosing this.
int state[100005];
void solve(int n, int p) {
for (auto i:adj[n]) {
if (i!=p) solve(i,n);
}
for (int i = 0; i < 2; i++) {
//suppose the endstate is i.
for (int j = 0; j < 2; j++) {
//j=0 means notake,
//j=1 means take.
//i^j is the state after considering all children.
//dp[c][j][x] where sum of x mod2 is i^j. if notake
//dp[c][j][x] where sum of x mod 2 is i^j if take.
//we first consider taking all =0 first.
vector<int> diff;
int total=0;
for (auto x:adj[n]) if (x!=p) {
total += dp[x][j][0];
diff.push_back(dp[x][j][1]-dp[x][j][0]);
}
sort(diff.begin(), diff.end());
if (i^j^state[n]==1 && diff.size()==0) {dp[n][i][j]=2e9; continue;}
else if (diff.size()==0) {dp[n][i][j]=total+j; continue;}
int start=0;
if (i^j^state[n]==1) {
total += diff[0];
start=1;
}
int ans=total;
while(start < diff.size()-1){
total += diff[start];
total += diff[start+1];
ans=min(ans,total);
start+=2;
}
dp[n][i][j]=ans+j;
}
}
}
signed main() {
int n; cin >> n;
for (int i = 0; i < n-1; i++) {
int a,b; cin >> a>> b;
a--;b--;
adj[a].push_back(b);
adj[b].push_back(a);
}
for (int i = 0; i < n; i++) cin >> state[i];
solve(0,-1);
int ans=min(dp[0][0][0], dp[0][0][1]);
if (ans>2e6) cout << "impossible";
else cout << ans;
}
