#include using namespace std; #define ll long long #define ull unsigned long long #define ld long double using LL = long long; using ULL = unsigned long long; using VI = vector; using VVI = vector; using VVVI = vector; using VL = vector; using VVL = vector; using VVVL = vector; using VB = vector; using VVB = vector; using VVVB = vector; using VD = vector; using VVD = vector; using VVVD = vector; using VC = vector; using VS = vector; using VVC = vector; using PII = pair; using PLL = pair; using PDD = pair; using PIL = pair; using MII = map; using MLL = map; using SI = set; using SL = set; using MSI = multiset; using MSL = multiset; template using MAXPQ = priority_queue; template using MINPQ = priority_queue< T, vector, greater >; const ll MOD = 1000000007; const ll MOD2 = 998244353; const ll INF = 1LL << 60; #define PI 3.14159265358979323846 #define FOR(i, a, b) for(int i = (a); i < (b); ++i) #define REP(i, n) FOR(i, 0, n) #define EACH(e, v) for(auto &e : v) #define RITR(it, v) for(auto it = (v).rbegin(); it != (v).rend(); ++it) #define ALL(v) v.begin(),v.end() vector x8={1,1,1,0,0,-1,-1,-1},y8={1,0,-1,1,-1,1,0,-1}; int dx4[4]={1,-1,0,0}, dy4[4]={0,0,1,-1}; /* memo -uf,RMQ(segtree),BIT,BIT2,SegTree,SegTreeLazy -isprime,Eratosthenes,gcdlcm,factorize,divisors,modpow,moddiv nCr(+modnCr,inverse,extend_euclid.powmod),tobaseB,tobase10 -dijkstra,Floyd,bellmanford,sccd,topological,treediamiter -compress1,compress2,rotate90 -co,ci,fo1,fo2,fo3,fo4 -bitsearch,binaryserach -bfs -SegTreedec,SegTreeLazydec */ void solve(){ ll N,M,K; cin >> N >> M >> K; vector> G(N); for(ll i = 0; i < M; i++){ ll u,v; cin >> u >> v; u--,v--; G[u].insert(v); G[v].insert(u); } VL b(N); for(ll i = 0; i < N; i++) cin >> b[i]; queue q; set vis; for(ll i = 0; i < N; i++){ if(G[i].size() == 1){ q.push(i); vis.insert(i); } } VL cnt(N); while(q.size()){ ll pos = q.front();q.pop(); for(auto to : G[pos]){ G[to].erase(pos); if(G[to].size() == 1 and !vis.count(to)){ vis.insert(to); q.push(to); } ll need = 0; if(cnt[pos] <= b[pos]) need = b[pos] - cnt[pos]; else need = b[pos] + K - cnt[pos]; cnt[pos] = (cnt[pos]+need) % K; cnt[to] = (cnt[to]+need) % K; } G[pos].clear(); } //for(ll i = 0; i < N; i++) cout << cnt[i] <<" "; //cout << '\n'; if(cnt == b) cout << "Yes" << '\n'; else cout << "No" << '\n'; } int main(){ cin.tie(0); ios_base::sync_with_stdio(0); int t; cin >> t; while(t--){ solve(); } }