#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int ll
const int inf = 2e15;
using pii = pair<int, int>;
#define fr first
#define sc second
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, m, k;
cin >> n >> m >> k;
vector<vector<int>> vec(n);
vector<bool> stop(n);
for (int i = 0; i < m; i ++) {
int u, v;
cin >> u >> v;
u --, v --;
vec[u].push_back(v);
vec[v].push_back(u);
}
int nrStop = 0;
for (int i = 0; i < n; i ++) {
char ch;
cin >> ch;
stop[i] = ch - '0';
nrStop += stop[i];
}
vector<int> start(k);
for (int i = 0; i < k; i ++) {
cin >> start[i];
start[i] --;
}
vector<int> distStart(n, inf), dp(n, 0);
distStart[start[0]] = 0;
queue<int> st;
st.push(start[0]);
vector<bool> viz(n, false);
while (!st.empty()) {
int x = st.front();
st.pop();
if ((x != start[0] && stop[x]) || viz[x])
continue;
viz[x] = true;
for (auto i : vec[x]) {
distStart[i] = min(distStart[i], 1 + distStart[x]);
st.push(i);
}
}
vector<int> ans(n, inf);
vector<bool> doubleStop(n, false);
for (int i = 0; i < n; i ++) {
for (auto j : vec[i])
if (stop[i] && stop[j])
doubleStop[i] = true;
}
vector<int> init(n, inf);
for (int i = 0; i < n; i ++)
if (stop[i]) {
st.push(i);
init[i] = 0;
}
viz.assign(n, 0);
while (!st.empty()) {
int x = st.front();
st.pop();
if (viz[x])
continue;
viz[x] = true;
for (auto i : vec[x]) {
init[i] = min(init[i], init[x] + 1);
st.push(i);
}
}
for (auto &i : init)
if (i == 0)
i += 2;
deque<int> dq;
vector<int> init2(n, inf);
for (int i = 0; i < n; i ++)
if (doubleStop[i]) {
init2[i] = 0;
dq.push_back(i);
}
viz.assign(n, 0);
while (!dq.empty()) {
int x = dq.front();
dq.pop_front();
if (viz[x])
continue;
viz[x] = true;
for (auto i : vec[x]) {
int cost = (stop[x] == 0);
init2[i] = min(init2[i], init2[x] + cost);
if (cost)
dq.push_back(i);
else
dq.push_front(i);
}
}
vector<int> deLa2La1(n);
for (int i = 0; i < n; i ++) {
deLa2La1[i] = init2[i] - init[i] + 2;
}
for (int rep = 0; rep <= nrStop; rep ++) {
int sum = 0;
for (int i = 1; i < k; i ++)
if (rep != 0)
sum += min(init[start[i]] + 2 * (rep - 1), init2[start[i]] + rep);
for (int i = 0; i < n; i ++)
ans[i] = min(ans[i], distStart[i] + sum);
//recalc distStart
auto distStart2 = distStart;
priority_queue<pii> pq;
for (int i = 0; i < n; i ++)
pq.push({-distStart[i], i});
viz.assign(n, 0);
while (!pq.empty()) {
auto x = pq.top();
pq.pop();
if (viz[x.sc])
continue;
viz[x.sc] = true;
x.fr *= -1;
distStart2[x.sc] = x.fr;
for (auto i : vec[x.sc]) {
int nw = 1;
if (stop[x.sc])
nw += distStart[x.sc];
else
nw += distStart2[x.sc];
if (distStart2[i] > nw) {
distStart2[i] = nw;
pq.push({-nw, i});
}
}
}
distStart = distStart2;
}
for (int i = 0; i < n; i ++)
cout << ans[i] << '\n';
}
/*
6 9 2
1 2
2 3
2 4
2 5
3 6
3 4
1 4
4 5
1 5
111100
5 6
am 2 variante
ma duc la cel mai apropiat stop:
acum trebuie sa vad astfel:
pentru varianta in care fac 2 de fiecare data ma costa init + 2 * (nrture - 1) = init2+ 1 * (nrture - tureFacute);
nrture = init2 - tureFacute - init + 2
vrem init2 - tureFacute minim
si init minim
.......X....XX
...X
*/
