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#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPD(i,a,b) for (int i = (a); i >= (b); --i)
#define FORI(i,n) REP(i,1,n)
#define FOR(i,n) REP(i,0,int(n)-1)
#define mp make_pair
#define pb push_back
#define sz(x) int((x).size())
#define dbg(v) cerr << #v << " = " << (v) << endl;
#define FOREACH(i,t) for (__typeof(t.begin()) i = t.begin(); i != t.end(); ++i)
#define fi first
#define se second
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
const ll INF = 10000000000000LL;

const int MAXN = 500100;

vector<int> g[MAXN];
int p[MAXN];
int a[MAXN];
int b[MAXN];
int n,m;
int r;
int lvl[MAXN];
vector<int> lvlV[MAXN];
int maxLvl;


void rootTree(int root) {
    r = root;
    p[r] = -1;
    lvl[r] = 0;
    maxLvl = 0;
    lvlV[0].pb(r);
    queue<int> q;
    q.push(r);
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        FOR(i, g[v].size()) {
            int v2 = g[v][i];
            if (v2 == p[v])
                continue;
            p[v2] = v;
            lvl[v2] = lvl[v] + 1;
            if (lvl[v2] > maxLvl)
                maxLvl = lvl[v2];
            lvlV[lvl[v2]].pb(v2);
            q.push(v2);
        }
    }
}

bool isLeaf(int v) {
    return g[v].size() == 1;
}

int main() {
    scanf("%d %d", &n, &m);
    FOR(i,n-1) {
        int x,y;
        scanf("%d %d", &x, &y);
        x--;
        y--;
        g[x].pb(y);
        g[y].pb(x);
    }
    FOR(i,m) {
        int x;
        scanf("%d", &x);
        a[i] = x;
        b[i] = x;
    }
    //special case
    if (m==n && n==2) {
        int res = a[0]-a[1];
        if (res < 0)
            res = -1*res;
        printf("%d\n", res);
        return 0;
    }
    rootTree(m);

    ll result = 0;
    for(int i=maxLvl;i>=0;i--) {
        FOR(j, lvlV[i].size()) {
            int v = lvlV[i][j];
            if (isLeaf(v))
                continue;

            //consider children of v -> make interval problem
            vector<int> children;
            FOR(k, g[v].size()) {
                int v2 = g[v][k];
                if (p[v] != v2) {
                    children.pb(v2);
                }
            }

            //solve Interval problem
            vector<pair<int, bool> > endPoints; //x, start=false
            ll d=0;
            FOR(k, children.size()) {
                int v2 = children[k];
                endPoints.pb(mp(a[v2],false));
                endPoints.pb(mp(b[v2],true));
                d += a[v2];
            }

            sort(endPoints.begin(), endPoints.end());

            int pos = 0;
            int left = 0;
            int right = children.size();
            ll bestD = INF;
            int bestA,bestB;
            bestA = bestB = 0;
            FOR(k, endPoints.size()) {
                pair<int, bool> event = endPoints[k];
                int x = event.first;
                bool ending = event.second;
                ll dx = x-pos;
                d += (left-right)*dx;
                pos = x;
                if (d<bestD) {
                    bestD = d;
                    bestA = bestB = pos;
                }
                if (d==bestD) {
                    bestB = pos;
                }
                if (ending)
                    left++;
                else
                    right--;
            }
            a[v] = bestA;
            b[v] = bestB;
            result += bestD;
        }

    }
    printf("%lld\n", result);
    return 0;
}