#include<bits/stdc++.h>
// #include<atcoder/all>
// #include<boost/multiprecision/cpp_int.hpp>

using namespace std;
// using namespace atcoder;
// using bint = boost::multiprecision::cpp_int;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<ll,ll>;
using vi = vector<ll>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ve = vector<vector<int>>;
using vb = vector<bool>;
using vvb = vector<vb>;
#define rep(i,n) for(ll i = 0;i < (ll)n;i++)
#define ALL(x) (x).begin(),(x).end()
#define sz(c) ((ll)(c).size())
#define LB(A,x) (int)(lower_bound(A.begin(),A.end(),x)-A.begin())
#define UB(A,x) (int)(upper_bound(A.begin(),A.end(),x)-A.begin())
// #define MOD 1000000007
#define MOD 998244353
template<typename T>using min_priority_queue=priority_queue<T,vector<T>,greater<T>>;
template<typename T>ostream&operator<<(ostream&os,vector<T>&v){for(int i = 0;i < v.size();i++)os<<v[i]<<(i+1!=v.size()?" ":"");return os;}
template<typename T>istream&operator>>(istream&is,vector<T>&v){for(T&in:v)is>>in;return is;}
template<typename T1,typename T2>ostream&operator<<(ostream&os,pair<T1,T2>&p){os<<p.first<<" "<<p.second;return os;}
template<typename T1,typename T2>istream&operator>>(istream&is,pair<T1,T2>&p){is>>p.first>>p.second;return is;}
template<typename T> inline bool chmax(T &a,T b){if(a < b){a = b;return true;}return false;}
template<typename T> inline bool chmin(T &a,T b){if(a > b){a = b;return true;}return false;}
ld dist(ld x1,ld y1,ld x2, ld y2){return sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}


// verifiy
// 重心分解 https://atcoder.jp/contests/abc291/submissions/54751516
// 分割統治 https://atcoder.jp/contests/ddcc2016-qual/submissions/54751335
struct edge{
  int to;
  // 情報を追加
  // ll we;
  // edge(int a,ll b):to(a),we(b){}
  edge(int k):to(k){}
};
using vve = vector<vector<edge>>;
struct CD{
  int n;
  vector<vector<edge>> v;
  vector<bool> used_centroid;
  vector<int> sub_sz;//内部用。重心分解の各段階で計算する部分木の大きさ
  vector<int> par;//重心分解後の親の頂点
  CD(vector<vector<edge>> &a) : 
        n((int)a.size()),v(a),used_centroid(a.size(),false),sub_sz(a.size(),1),par(a.size(),-1){}
  int subtree_sz(int ov,int pre){
    int res = 1;
    for(int i = 0;i < (int)v[ov].size();i++){
      int to = v[ov][i].to;
      if(to == pre || used_centroid[to])continue;
      res += subtree_sz(to,ov);
    }
    sub_sz[ov] = res;
    return res;
  }
  int get_centroid(int ov,int now_sz){
    while(1){
      int mx = -1,can_v = -1;
      for(auto nv : v[ov]){
        // 重心でない / 逆走しない 条件
        if(used_centroid[nv.to] || sub_sz[nv.to] > sub_sz[ov])continue;
        mx = max(mx,sub_sz[nv.to]);
        if(mx == sub_sz[nv.to])can_v = nv.to;
      }
      if(mx <= now_sz/2)break;
      ov = can_v;
    }
    return ov;
  }
  // ovが所属する部分木に対する重心分解 
  // 使うときはpre = -1を入れる
  // O(nlogn)
  void cd(int ov,int pre,string &s,ll &res){//分割統治に必要なら引数を増やす
    subtree_sz(ov,-1);
    ov = get_centroid(ov,sub_sz[ov]);
    par[ov] = pre;
    used_centroid[ov] = true;
    vi ds;ds.emplace_back(0);
    // /*
    // 重心ovに対して➀u->ov->v ➁ov->u となる頂点対に対する部分問題を書く。必要ならば下のdfsを用いる
    auto count = [](vi &d,string &s,int ov){
      sort(ALL(d));
      ll res = 0;
      rep(i,sz(d)){
        int k = LB(d,-((s[ov] == '1' ? 1 : -1) + d[i])+1);
        res += sz(d)-k;
      }
      return res;
    };
    rep(i,sz(v[ov])){
      int to = v[ov][i].to;
      if(used_centroid[to])continue;
      vi dt;
      auto dfs = [&](auto &dfs,int OV,int PRE,ll cnt)->void{
        dt.emplace_back(cnt + (s[OV] == '1' ? 1 : -1));
        for(auto nv : v[OV]){
          if(nv.to == PRE || used_centroid[nv.to])continue;
          dfs(dfs,nv.to,OV,cnt + (s[OV] == '1' ? 1 : -1));
        }
      };
      dfs(dfs,v[ov][i].to,ov,0);
      ll we = count(dt,s,ov);
      res -= we;
      ds.insert(ds.end(),ALL(dt));
    }
    ll we = count(ds,s,ov);
    res += we;
    if(s[ov] == '1')res++;
    for(int i = 0;i < (int)v[ov].size();i++){
      if(used_centroid[v[ov][i].to])continue;
      cd(v[ov][i].to,ov,s,res);
    }
  }
};

int main(){
  
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  vve v(n);
  rep(i,n-1){
    int a,b;cin >> a >> b;
    a--;b--;
    v[a].emplace_back((edge)(b));
    v[b].emplace_back((edge)(a));
  }
  CD cd(v);
  string s;cin >> s;
  ll res = 0;
  cd.cd(0,-1,s,res);
  res /= 2;
  cout << res << "\n";
  

  return 0;
}