#include <cstdio>#include <cstring>typedef long long ll;using namespace std;const int N = 1500;const int MOD = 1e9 + 7;int Scan() {    int x = 0, f = 1; char ch = getchar();    while (ch < ‘0‘ || ch > ‘9‘) { if (ch == ‘-‘) f = -1; ch = getchar(); }    while (ch >= ‘0‘ && ch <= ‘9‘) { x = (x << 3) + (x << 1) + ch - ‘0‘; ch = getchar(); }    return x * f;}struct Edge {    int to, next;} edge[N*2];int head[N], cntE;void addedge(int u, int v) {    edge[cntE].to = v; edge[cntE].next = head[u]; head[u] = cntE++;    edge[cntE].to = u; edge[cntE].next = head[v]; head[v] = cntE++;}int a[N], n, m;int sz[N];      // sz[u] 记录 根为 u 点 有多少个异或值int val[N][N];  // val[u][i] 记录 u 点 第 i 个的异或值都是什么int dp[N][N];   // dp[u][i] 记录 根为 u 点 异或值为i的数量int vis[N][N];  // vis[i] 记录 i 这个 异或值是否出现过int tmp[N];void add(int &a, int b) {    a = ((ll)a + b) % MOD;}void dfs(int u, int fa) {    dp[u][ a[u] ] = 1;    val[u][ sz[u]++ ] = a[u];    vis[u][ a[u] ] = true;    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].to;        if (v == fa) continue;        dfs(v, u);        memset(tmp, 0, sizeof(int)*m);        for (int j = sz[u]-1; j >= 0; --j) {            for (int k = 0; k < sz[v]; ++k) {                int a = val[u][j], b = val[v][k];                add(tmp[a^b], (ll)dp[u][a] * dp[v][b] % MOD);                if (!vis[u][a^b]) val[u][sz[u]++] = a^b, vis[u][a^b] = true;            }        }        for (int j = 0; j < m; ++j) add(dp[u][j], tmp[j]);    }}void init() {    memset(head, -1, sizeof(int)*(n+2));    memset(sz, 0, sizeof(int)*(n+2));    for (int i = 0; i <= n; ++i)        memset(dp[i], 0, sizeof(int)*m), memset(vis[i], 0, sizeof(int)*m);    cntE = 0;}int main(){    //freopen("in.txt", "r", stdin);    int T = Scan();    while (T--) {        n = Scan(), m = Scan();        init();        for (int i = 1; i <= n; ++i) a[i] = Scan();        for (int i = 1; i < n; ++i) addedge(Scan(), Scan());        dfs(1, 0);        for (int k = 0; k < m; ++k) {            int ans = 0;            for (int i = 1; i <= n; ++i) add(ans, dp[i][k]);            printf("%d%c", ans, k==m-1?‘\n‘:‘ ‘);        }    }    return 0;}