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

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 $\mathrm{nn\; towns.}$

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 

题解：商品的个数最多只有100，所以枚举商品。用BFS构造最短路，注意的是起始距离变成了1。

#pragma warning(disable:4996)
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define mem(arr, in) memset(arr, in, sizeof(arr))
using namespace std;

const int maxn = 100205;
const int INF = 1e9 + 7;

int n, m, k, s;
int use[maxn], d[maxn][105], co[maxn];

vector<int> v[maxn];

void minroad(int st) {
    queue<int> q;
    q.push(st + n);
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = 0; i < v[u].size(); i++) {
            int to = v[u][i];
            if (!d[to][st]) {
                d[to][st] = d[u][st] + 1;
                q.push(to);
            }
        }
    }
}

int main()
{
    while (scanf("%d %d %d %d", &n, &m, &k, &s) != EOF) {
        for (int i = 1; i <= n; i++) scanf("%d", co + i), v[co[i] + n].push_back(i);
        for (int i = 1; i <= m; i++) {
            int a, b;
            scanf("%d%d", &a, &b);
            v[a].push_back(b);
            v[b].push_back(a);
        }
        for (int i = 1; i <= k; i++) minroad(i);
        for (int i = 1; i <= n; i++) {
            sort(d[i] + 1, d[i] + k + 1);
            int ans = 0;
            for (int j = 1; j <= s; j++) ans += d[i][j];
            printf("%d ", ans - s);
        }
        printf("
");
    }
    return 0;
}