Problem:
Score : 1500 points

Problem Statement

You are given a tree with N vertices numbered 1, 2, ..., N. The i-th edge connects Vertex a_i and Vertex b_i.
You are also given a string S of length N consisting of 0 and 1. The i-th character of S represents the number of pieces placed on Vertex i.

Snuke will perform the following operation some number of times:

Choose two pieces the distance between which is at least 2, and bring these pieces closer to each other by 1. More formally, choose two vertices u and v, each with one or more pieces, and consider the shortest path between them. Here the path must contain at least two edges. Then, move one piece from u to its adjacent vertex on the path, and move one piece from v to its adjacent vertex on the path.

By repeating this operation, Snuke wants to have all the pieces on the same vertex. Is this possible?
If the answer is yes, also find the minimum number of operations required to achieve it.

Constraints

2 \leq N \leq 2000

|S| = N

S consists of 0 and 1, and contains at least one 1.

1 \leq a_i, b_i \leq N(a_i \neq b_i)

The edges (a_1, b_1), (a_2, b_2), ..., (a_{N - 1}, b_{N - 1}) forms a tree.

Input

Input is given from Standard Input in the following format:

N
S
a_1 b_1
a_2 b_2
:
a_{N - 1} b_{N - 1}

Output

If it is impossible to have all the pieces on the same vertex, print -1. If it is possible, print the minimum number of operations required.

Sample Input 1

7
0010101
1 2
2 3
1 4
4 5
1 6
6 7

Sample Output 1

3

We can gather all the pieces in three operations as follows:

Choose the pieces on Vertex 3 and 5.

Choose the pieces on Vertex 2 and 7.

Choose the pieces on Vertex 4 and 6.

Sample Input 2

7
0010110
1 2
2 3
1 4
4 5
1 6
6 7

Sample Output 2

-1

Sample Input 3

2
01
1 2

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//#include <bits/stdc++.h>
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
//#include "boost/multiprecision/cpp_int.hpp"
//typedef boost::multiprecision::cpp_int LL;
typedef long double dd;
//#define i_7 (ll)(1E9+7)
//#define i_7 998244353
ll i_7;
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
typedef pair<dd,dd> d_d;
ll inf=(ll)1E16;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fi first
#define se second
#define endl "\n"
#define SORT(v) sort(v.begin(),v.end())
#define ERASE(v) v.erase(unique(v.begin(),v.end()),v.end())
#define POSL(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define POSU(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}


void mod_print(ll k){
    ll P=1000;
    rep(y,1,P){
        ll x=mod(y*k);
        if(x+P>=i_7){
            x-=i_7;
        }
        if(abs(x)<=P){
            cout<<x<<"/"<<y<<endl;
            return;
        }
    }
    cout<<"nun"<<endl;
}
//////////////////////////

#define N 3000
ll n;
string s;
vector<ll>v[N];

ll kaz[N];
ll dp[N];//min
ll ep[N];//max
void init(){
    memset(kaz,0,sizeof(kaz));
    memset(dp,0,sizeof(dp));
    memset(ep,0,sizeof(ep));
}
void dfs(ll id,ll par){
    if(s[id]=='1'){
        kaz[id]++;
    }
    ll sum=0;
    for(auto x:v[id]){
        if(x==par)continue;
        dfs(x,id);
        kaz[id]+=kaz[x];
        ep[id]+=ep[x]+kaz[x];
        sum+=(ep[x]+kaz[x]);
    }
    dp[id]=ep[id];
    bool f=true;
    
    for(auto x:v[id]){
        if(x==par)continue;
        if(sum-(ep[x]+kaz[x])<(dp[x]+kaz[x])){
            f=false;
            chmin(dp[id],dp[x]+ep[x]+2*kaz[x]-sum);
        }
    }
    if(f){
        dp[id]=ep[id]%2;
    }
}

int main(){fastio
    cin>>n>>s;
    rep(i_,1,n-1){
        ll a,b;cin>>a>>b;
        a--;b--;
        v[a].pb(b);
        v[b].pb(a);
    }
    ll ans=inf;
    rep(i,0,n-1){
        init();
        dfs(i,-1);
        
        if(dp[i]==0){
            chmin(ans,ep[i]/2);
            //cout<<i<<":"<<ep[i]<<endl;
        }
        /*
        if(i==5)rep(j,0,n-1){
            cout<<j<<":"<<dp[j]<<" "<<ep[j]<<endl;
        }*/
    }
    if(ans==inf)ans=-1;
    cout<<ans<<endl;
    return 0;
}