#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const ll INFLL = 1LL<<60;
const ll MOD = 998244353;
const double INFD = 1.0E10;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, -1, 0, 1};
//const int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
//const int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using Pair = pair<ll, ll>;
using Graph = vector<vector<ll>>;
using mint = atcoder::modint998244353;


int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
    
    
    int T; cin >> T;
    while (T--){
        ll n, m, k; cin >> n >> m >> k;
        Graph G(n);
        for (int i = 0; i < m; i++){
            int u, v; cin >> u >> v;
            u--; v--;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        vector<ll> b(n);
        for (int i = 0; i < n; i++) cin >> b[i];
        
        
        if (m == n - 1){
            //木のとき、葉から見ていく
            auto rec = [&](auto &&self, int v, int p = -1) -> int{
                int ret = 0;
                for (auto nv : G[v]){
                    if (nv == p) continue;
                    ret = (ret + self(self, nv, v)) % k;
                }
                return (b[v] + k - ret) % k;
            };
            
            auto ret = rec(rec, 0);
            cout << (ret == 0 ? "Yes" : "No") << '\n';
        }
        else{
            //必ず閉路が存在する
            if (k % 2){
                //このときは必ずいけるはず
                cout << "Yes" << '\n';
            }
            else{
                //偶数の場合は偶奇が合えばいけるんじゃね？
                int total = 0;
                for (int i = 0; i < n; i++) total = (total + b[i]) % k;
                cout << (total % 2 == 0? "Yes" : "No") << '\n';
            }
        }
    }
    
    
    return 0;
}