#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = (int)1e5 + 1, md = (int)1e9 + 7;
vector<int> pw(N, 1);
ll MD(ll x) {
return (x % md + md) % md;
}
int f(bool x, bool y, int a, int b, int c) {
if (~a) {
if (!a) return f(x, y, -1, b, c);
return MD((ll)pw[a - 1] * MD(f(x, y, -1, b, c) + f(!x, !y, -1, b, c)));
}
else if (~b) return MD((ll)(b ? pw[b - 1] : x) * f(true, y, -1, -1, c));
else return (c ? pw[c - 1] : y);
}
int main() {
ios::sync_with_stdio(false); cin.tie(nullptr);
for (int i = 1; i < N; i++)
pw[i] = MD(pw[i - 1] << 1);
int n;
cin >> n;
vector<int> a(n);
for (auto &i : a)
cin >> i;
int q;
cin >> q;
int l[q], r[q], X[q];
for (int i = 0; i < q; i++) {
cin >> l[i] >> r[i] >> X[i]; l[i]--; r[i]--;
}
int ans = 0;
for (int x = 0; x < 8; x++) {
for (int y = 0; y < 8; y++) {
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
int A = 0, B = 0, C = 0;
for (int k = 0; k < q; k++) {
bool tr1 = (X[k] >> x) & 1, tr2 = (X[k] >> y) & 1;
bool ok1 = (l[k] <= i && i <= r[k]), ok2 = (l[k] <= j && j <= r[k]);
A += (tr1 && tr2 && ok1 && ok2);
B += (tr1 && ok1 && (!tr2 || !ok2));
C += (tr2 && ok2 && (!tr1 || !ok1));
}
int z = 1 + (i != j);
int vl = MD((ll)z * pw[x] * pw[y]);
int pr = MD((ll)(i + 1) * (n - j));
int cnt = f((a[i] >> x) & 1, (a[j] >> y) & 1, A, B, C);
ans = MD(ans + MD(MD(MD(vl * pr) * cnt) * pw[q - A - B - C]));
}
}
}
}
cout << ans;
}
