Problem:
Score : 800 points

Problem Statement

There is a tree with N vertices numbered 1 through N.
The i-th edge connects Vertex x_i and y_i.
Each vertex is painted white or black.
The initial color of Vertex i is represented by a letter c_i.
c_i = W represents the vertex is white; c_i = B represents the vertex is black.

A cat will walk along this tree.
More specifically, she performs one of the following in one second repeatedly:

Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.

Invert the color of the vertex where she is currently.

The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex.
At least how many seconds does it takes for the cat to achieve her objective?

Constraints

1 ≤ N ≤ 10^5

1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)

The given graph is a tree.

c_i = W or c_i = B.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N

Output

Print the minimum number of seconds required to achieve the objective.

Sample Input 1

5
1 2
2 3
2 4
4 5
WBBWW

Sample Output 1

5

The objective can be achieved in five seconds, for example, as follows:

Start at Vertex 1. Change the color of Vertex 1 to black.

Move to Vertex 2, then change the color of Vertex 2 to white.

Change the color of Vertex 2 to black.

Move to Vertex 4, then change the color of Vertex 4 to black.

Move to Vertex 5, then change the color of Vertex 5 to black.

Sample Input 2

6
3 1
4 5
2 6
6 1
3 4
WWBWBB

Sample Output 2

7

Sample Input 3

1
B

Sample Output 3

0

Sample Input 4

20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW

Sample Output 4

21

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

/*

This Submission is to determine how many 120/240 min const. delivery point there are.

//info
120 req. steps <= 5
*/
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
#include <complex>
#include <cctype>
#include <bitset>
using namespace std;
typedef string::const_iterator State;
#define Ma_PI 3.141592653589793
#define eps 1e-5
#define LONG_INF 1e18
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007LL
#define MOD 998244353LL
#define seg_size 262144*4
#define REP(a,b) for(long long a = 0;a < b;++a)

unsigned long xor128() {
	static unsigned long x = time(NULL), y = 362436069, z = 521288629, w = 88675123;
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
double dot(complex<double> a, complex<double> b) {
	return a.real() * b.real() + a.imag() * b.imag();
}
double gyaku_dot(complex<double> a, complex<double> b) {
	return a.real() * b.imag() - a.imag() * b.real();
}
double leng(complex<double> a) {
	return sqrt(a.real() * a.real() + a.imag() * a.imag());
}
double angles(complex<double> a, complex<double> b) {
	double cosine = dot(a, b) / (leng(a) * leng(b));
	double sine = gyaku_dot(a, b) / (leng(a) * leng(b));
	double kaku = acos(min((double)1.0, max((double)-1.0, cosine)));
	if (sine <= 0) {
		kaku = 2 * Ma_PI - kaku;
	}
	return kaku;
}
vector<int> convex_hull(vector<complex<double>> a) {
	vector<int> ans;
	double now_minnest = a[0].real();
	int now_itr = 0;
	REP(i, a.size()) {
		if (now_minnest > a[i].real()) {
			now_minnest = a[i].real();
			now_itr = i;
		}
	}
	ans.push_back(now_itr);
	complex<double> ba(0, 1);
	while (true) {
		int now_go = 0;
		double now_min = 0;
		double now_length = 0;
		int starter = ans[ans.size() - 1];
		for (int i = 0; i < a.size(); ++i) {
			if (i != starter) {
				double goa = angles(ba, a[i] - a[starter]);
				if (goa - now_min >= eps || (abs(goa - now_min) <= eps && (abs(a[i] - a[starter]) - now_length) >= eps)) {
					now_min = goa;
					now_go = i;
					now_length = abs(a[i] - a[starter]);
				}
			}
		}
		if (now_go == ans[0]) break;
		ans.push_back(now_go);
		ba = complex<double>(a[now_go] - a[starter]);
	}
	return ans;
}
string s;
vector<int> vertexs[200000];
int cnt[200000];
int distances[200000];
void dfs(int now, int back) {
	REP(q, vertexs[now].size()) {
		if (vertexs[now][q] == back) continue;
		if (cnt[vertexs[now][q]] == 0) continue;
		distances[vertexs[now][q]] = distances[now];
		if (s[vertexs[now][q]] == 'W') distances[vertexs[now][q]] += 2;
		dfs(vertexs[now][q], now);
	}
}
int main() {
	iostream::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	REP(i, n - 1) {
		int a, b;
		cin >> a >> b;
		a--; b--;
		vertexs[a].push_back(b);
		vertexs[b].push_back(a);
		cnt[a]++;
		cnt[b]++;
	}
	cin >> s;
	int hoge = 0;
	for (int i = 0; i < s.length(); ++i) {
		if (s[i] == 'W')hoge++;
	}
	if (hoge <= 1) {
		cout << hoge << endl;
		return 0;
	}
	queue<int> next;
	for (int i = 0; i < n; ++i) {
		if (cnt[i] == 1&&s[i] == 'B') {
			next.push(i);
		}
	}
	while (next.empty() == false) {
		int now = next.front(); next.pop();
		cnt[now] = 0;
		REP(i, vertexs[now].size()) {
			cnt[vertexs[now][i]]--;
			if (s[vertexs[now][i]] == 'B'&&cnt[vertexs[now][i]] == 1) {
				next.push(vertexs[now][i]);
			}
		}
	}
	int ans = 0;
	for (int i = 0; i < n; ++i) {
		if (cnt[i] != 0) {
			int cnter = 0;
			if (s[i] == 'W')cnter++;
			REP(q, vertexs[i].size()) {
				if (cnt[vertexs[i][q]] != 0) {
					cnter++;
					ans++;
				}
			}
			ans += cnter % 2;
			if (cnter % 2 == 0) {
				s[i] = 'B';
			}
			else {
				s[i] = 'W';
			}
		}
	}
	REP(i, n) {
		if (cnt[i] != 0) {
			dfs(i, -1);
			break;
		}
	}
	int now_max = -1;
	int now_itr = 0;
	REP(i, n) {
		if (cnt[i] == 1) {
			if (now_max < distances[i]) {
				now_max = distances[i];
				now_itr = i;
			}
		}
		distances[i] = 0;
	}
	if (s[now_itr] == 'W') {
		distances[now_itr] = 2;
	}
	dfs(now_itr, -1);
	now_max = 0;
	REP(q, n) {
		if (cnt[q] != 0) {
			now_max = max(now_max, distances[q]);
		}
	}
	cout << ans-now_max << endl;
}