Fair

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

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

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 $\mathrm{nn towns.}$

Input

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

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

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

Output

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

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 $\mathrm{11 you\; can\; bring\; goods\; from\; towns 11 (00 coins), 22 (11 coin)\; and 44 (11 coin).\; Total\; numbers\; of\; coins\; is 22.}$

Town $\mathrm{22:\; Goods\; from\; towns 22 (00), 11 (11), 33 (11).\; Sum\; equals 22.}$

Town $\mathrm{33:\; Goods\; from\; towns 33 (00), 22 (11), 44 (11).\; Sum\; equals 22.}$

Town $\mathrm{44:\; Goods\; from\; towns 44 (00), 11 (11), 55 (11).\; Sum\; equals 22.}$

Town $\mathrm{55:\; Goods\; from\; towns 55 (00), 44 (11), 33 (22).\; Sum\; equals 33.}$

bfs，直接搜就好了。

#include <bits/stdc++.h>
#define maxn 200005
using namespace std;
struct edge
{
    int u=0,v=0,next=0;
}edge[maxn];
int head[maxn];
int a[maxn];
int cnt=0;
void addedge(int u,int v)
{
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
bool goods[105];
bool vis[maxn];
struct data
{
    int x=0,step=0;
};
int bfs(int s,int sum)
{
    memset(goods,false,sizeof(goods));
    memset(vis,false,sizeof(vis));
    queue<data> q;
    data now,nex;
    int val=0,i;
    now.x=s;
    goods[a[now.x]]=true;
    int all=1;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(all==sum) return val;
        for(i=head[now.x];i!=-1;i=edge[i].next)
        {
            nex.x=edge[i].v;
            nex.step=now.step+1;
            if(!goods[a[nex.x]])
            {
                all++;
                val+=nex.step;
                if(all==sum) return val;
                q.push(nex);
                goods[a[nex.x]]=true;
                vis[nex.x]=true;
            }
        }
        for(i=head[now.x];i!=-1;i=edge[i].next)
        {
            nex.x=edge[i].v;
            if(!vis[nex.x])
            {   //val+=nex.step;
                q.push(nex);
                vis[nex.x]=true;
            }
        }
    }
}
int main()
{
    int n,m,k,s,i;
    memset(head,-1,sizeof(head));
    cin>>n>>m>>k>>s;
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    int u,v;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&u,&v);
        addedge(u,v);
        addedge(v,u);
    }
    for(i=1;i<=n;i++)
    {   if(i!=1) printf(" ");
        printf("%d",bfs(i,s));
    }
    printf("
");
    return 0;
}