#include #include typedef long long int ll; typedef long long int ull; #define MP make_pair using namespace std; using namespace atcoder; typedef pair P; // const ll MOD = 998244353; const ll MOD = 1000000007; // using mint = modint998244353; using mint = modint1000000007; const double pi = 3.1415926536; const int MAX = 2000005; long long fac[MAX], finv[MAX], inv[MAX]; template using min_priority_queue = priority_queue, greater>; void COMinit() { fac[0] = fac[1] = 1; finv[0] = finv[1] = 1; inv[1] = 1; for (int i = 2; i < MAX; i++){ fac[i] = fac[i - 1] * i % MOD; inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD; finv[i] = finv[i - 1] * inv[i] % MOD; } } // 二項係数計算 long long COM(int n, int k){ if (n < k) return 0; if (n < 0 || k < 0) return 0; return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD; } ll gcd(ll x, ll y) { if (y == 0) return x; else if (y > x) { return gcd (y, x); } else return gcd(x % y, y); } ll lcm(ll x, ll y) { return x / gcd(x, y) * y; } ll my_sqrt(ll x) { ll m = 0; ll M = 3000000001; while (M - m > 1) { ll now = (M + m) / 2; if (now * now <= x) { m = now; } else { M = now; } } return m; } ll keta(ll n) { ll ret = 0; while (n) { n /= 10; ret++; } return ret; } ll ceil(ll n, ll m) { // n > 0, m > 0 ll ret = n / m; if (n % m) ret++; return ret; } ll pow_ll(ll x, ll n) { if (n == 0) return 1; if (n % 2) { return pow_ll(x, n - 1) * x; } else { ll tmp = pow_ll(x, n / 2); return tmp * tmp; } } vector compress(vector v) { // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] vector u = v; sort(u.begin(), u.end()); u.erase(unique(u.begin(),u.end()),u.end()); map mp; for (int i = 0; i < u.size(); i++) { mp[u[i]] = i; } for (int i = 0; i < v.size(); i++) { v[i] = mp[v[i]]; } return v; } vector

v[100001]; bool visited1[100001]; bool visited2[100001]; P dfs1(int cur, int par, int type, ll num, ll k) { // type1 x + num type2 num - x visited1[cur] = true; ll m, M; if (type == 1) { m = 1 - num; M = k - num; } else { m = num - k; M = num - 1; } for (auto x : v[cur]) { int nex = x.first; ll c = x.second; if (visited1[nex]) continue; P p = dfs1(nex, cur, 3 - type, c - num, k); m = max(p.first, m); M = min(p.second, M); } return P(m, M); } P dfs2(int cur, int par, int type, ll num, ll k) { // type1 x + num type2 num - x visited2[cur] = true; ll m, M; if (type == 1) { m = 1 - num; M = k - num; } else { m = num - k; M = num - 1; } for (auto x : v[cur]) { int nex = x.first; ll c = x.second; if (visited2[nex]) continue; P p = dfs2(nex, cur, 3 - type, c - num, k); m = max(p.first, m); M = min(p.second, M); } return P(m, M); } int main() { for (int i = 0; i <= 100000; i++) { visited1[i] = false; visited2[i] = false; } ll n, m, k; cin >> n >> m >> k; dsu d(n); for (int i = 1; i <= m; i++) { ll x, y, z; cin >> x >> y >> z; x--; y--; v[x].push_back({y, z}); v[y].push_back({x, z}); d.merge(x, y); } mint ans = 1; mint ans2 = 1; for (auto g : d.groups()) { P p1 = dfs1(g[0], -1, 1, 0, k); mint tmp1 = max(p1.second - p1.first + 1, (ll)0); P p2 = dfs2(g[0], -1, 1, 0, k - 1); mint tmp2 = max(p2.second - p2.first + 1, (ll)0); // cout << g[0] << ' ' << tmp1.val() << ' ' << tmp2.val() << endl; ans *= tmp1; ans2 *= tmp2; } mint ret = ans - ans2; cout << ret.val() << endl; return 0; }