D. Fair

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Some company is going to hold a fair in Byteland. There are nn towns in Byteland and mm two-way roads between towns. Of course, you can reach any town from any other town using roads.

There are kk types of goods produced in Byteland and every town produces only one type. To hold a fair you have to bring at least ssdifferent types of goods. It costs d(u,v)d(u,v) coins to bring goods from town uu to town vv where d(u,v)d(u,v) is the length of the shortest path from uu to vv. Length of a path is the number of roads in this path.

The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of nn towns.

Input

There are 44 integers nn, mm, kk, ss in the first line of input (1≤n≤1051≤n≤105, 0≤m≤1050≤m≤105, 1≤s≤k≤min(n,100)1≤s≤k≤min(n,100)) — the number of towns, the number of roads, the number of different types of goods, the number of different types of goods necessary to hold a fair.

In the next line there are nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k), where aiai is the type of goods produced in the ii-th town. It is guaranteed that all integers between 11 and kk occur at least once among integers aiai.

In the next mm lines roads are described. Each road is described by two integers uu vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — the towns connected by this road. It is guaranteed that there is no more than one road between every two towns. It is guaranteed that you can go from any town to any other town via roads.

Output

Print nn numbers, the ii-th of them is the minimum number of coins you need to spend on travel expenses to hold a fair in town ii. Separate numbers with spaces.

Examples

input

Copy

5 5 4 3
1 2 4 3 2
1 2
2 3
3 4
4 1
4 5

output

Copy

2 2 2 2 3 

input

Copy

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

output

Copy

1 1 1 2 2 1 1 

Note

Let's look at the first sample.

To hold a fair in town 11 you can bring goods from towns 11 (00 coins), 22 (11 coin) and 44 (11 coin). Total numbers of coins is 22.

Town 22: Goods from towns 22 (00), 11 (11), 33 (11). Sum equals 22.

Town 33: Goods from towns 33 (00), 22 (11), 44 (11). Sum equals 22.

Town 44: Goods from towns 44 (00), 11 (11), 55 (11). Sum equals 22.

Town 55: Goods from towns 55 (00), 44 (11), 33 (22). Sum equals 33.

#include <bits/stdc++.h>
#define ll long long 
using namespace std;
const int inf = 0x3f3f3f3f;
const int mn = 100010;
const double eps = 1e-9;

int cnt;
int to[2 * mn], nx[2 * mn], fr[mn];
void add_edge(int a, int b)
{
    to[++cnt] = b;
    nx[cnt] = fr[a];
    fr[a] = cnt;
}

int n;
vector<int> g[110];
int dis[mn][110]; 
void bfs(int x)
{
    for (int i = 1; i <= n; i++)
        dis[i][x] = inf;
    
    queue<int> q;
    for (int i = 0, sz = g[x].size(); i < sz; i++)
    {
        dis[g[x][i]][x] = 0;    // 到自身點距離為0
        q.push(g[x][i]);
    }
    
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        
        for (int i = fr[u]; i != -1; i = nx[i])
        {
            if (dis[to[i]][x] != inf)   // to[i]已計算過距離
                continue;
            dis[to[i]][x] = dis[u][x] + 1;
            q.push(to[i]);
        }
    }
}

int main()
{
    memset(fr, -1, sizeof fr);
    
    int m, k, s;
    scanf("%d %d %d %d", &n, &m, &k, &s);
    for (int i = 1; i <= n; i++)
    {
        int t;
        scanf("%d", &t);
        g[t].push_back(i);  // 顏色為t的點標號
    }
    
    for (int i = 1; i <= m; i++)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        add_edge(a, b);
        add_edge(b, a);
    }
    
    for (int i = 1; i <= k; i++)    // bfs顏色, 求該顏色到各點的最短距離
        bfs(i);
    
    for (int i = 1; i <= n; i++)
    {
        sort(dis[i] + 1, dis[i] + k + 1);
        
        int ans = 0;
        for (int j = 1; j <= s; j++)    // 取距離前s的顏色
            ans += dis[i][j];
        printf("%d ", ans);
    }
    printf("\n");
    
    return 0;
}