#include <fstream>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

ifstream fin("distante.in");
ofstream fout("distante.out");

vector<pair<int, int> > g[50010];
int n, m, s, t;
int distante[50010];

void citireGraf() {
    fin >> n >> m >> s;
    for(int i = 1; i <= n; i++)
        g[i].clear();
    for(int i = 1; i <= n; i++)
        fin >> distante[i];
    for(int i = 1; i <= m; i++) {
        int a, b, c;
        fin >> a >> b >> c;
        g[a].push_back({b, c});
        g[b].push_back({a, c});
    }
}

///dijkstra cu heap - 50p
/*void solve() {
    int d[50010];
    for(int i = 1; i <= n; i++)
        d[i] = -1;
    priority_queue<pair<int, int> > q;
    q.push({0, s});
    bool ok = true;
    while(!q.empty()) {
        int nod = q.top().second;
        int cost = q.top().first;
        q.pop();
        if(d[nod] != -1) continue;
        d[nod] = -cost;
        if(d[nod] != distante[nod]) {
            ok = false;
            break;
        }
        for(int i = 0; i < g[nod].size(); i++)
            q.push({cost - g[nod][i].second, g[nod][i].first});
    }

    if(ok) fout << "DA\n";
    else fout << "NU\n";
}*/

///solutia optima e O(n+m) pt ca nu trebuie sa calculam distantele, ci doar sa verificam daca sunt corecte
void solve() {
    bool ok = true;
    if(distante[s]) {
        fout << "NU\n";
        return;
    }
    for(int i = 1; i <= n; i++) {
        if(i == s) continue;
        bool egal = false;
        for(int j = 0; j < g[i].size(); j++) {
            int vecin = g[i][j].first;
            int c = g[i][j].second;
            if(distante[i] > distante[vecin]+c) {
                fout << "NU\n";
                return;
            } else if(distante[i] == distante[vecin]+c)
                egal = true;
        }
        if (!egal) {
            fout << "NU\n";
            return;
        }
    }
    fout << "DA\n";
}


int main() {
    fin >> t;
    while(t--) {
        citireGraf();
        solve();
    }
}