#include const int maxn = 2e5 + 3, maxa = (1 << 10); const long long mo = 1e9 + 7; int n, m, a[maxn], b[maxn], K; long long ac[2][maxa], bc[2][maxa], cac[maxa], cbc[maxa]; int main() { scanf("%d%d%d", &n, &m, &K); for(int i = 0; i < n; ++i) { scanf("%d", a + i); } for(int i = 0; i < m; ++i) { scanf("%d", b + i); } for(int i = 0; i < n; ++i) { for(int j = 0; j < maxa; ++j) { int u = j ^ a[i]; ac[i & 1][u] = ac[~i & 1][j]; } ac[i & 1][a[i]] += 1; if(ac[i & 1][a[i]] >= mo) ac[i & 1][a[i]] -= mo; for(int j = 0; j < maxa; ++j) { cac[j] = (cac[j] + ac[i & 1][j]) % mo; } } for(int i = 0; i < m; ++i) { for(int j = 0; j < maxa; ++j) { int u = j ^ b[i]; bc[i & 1][u] = bc[~i & 1][j]; } bc[i & 1][b[i]] += 1; if(bc[i & 1][b[i]] >= mo) bc[i & 1][b[i]] -= mo; for(int j = 0; j < maxa; ++j) { cbc[j] = (cbc[j] + bc[i & 1][j]) % mo; } } long long ans = 0; for(int i = 0; i < maxa; ++i) { ans = (ans + cac[i] * (long long)cbc[i ^ K]) % mo; } printf("%lld\n", (ans + (long long)mo) % mo); return 0; }