#include <iostream>

using namespace std;

const int N = 400, MOD = 1000000007;

int n, m, x, v[N], a[N][N], ans;

int main() {
    scanf("%d%d%d", &n, &m, &x);
    for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
    // 构建初始xor
    for (int i = 0; i < 30; ++i) { // 每一位都是独立的
        for (int j = 1; j <= n; ++j) {
            a[i + 1][j] = v[j] >> i & 1;
        }
        a[i + 1][n + 1] = x >> i & 1;
    }
    for (int i = 1, op, l, r; i <= m; ++i) {
        scanf("%d%d%d", &op, &l, &r);
        for (int j = 1; j <= n; ++j) { // l~r每个数xor起来是op
            a[30 + i][j] = l <= j && j <= r;
        }
        a[30 + i][n + 1] = op;
    }

    int r = 1, z = m + 30, cnt = 0;
    // 找第一个有1的行
    for (int c = 1; c <= n; ++c) {
        for (int i = r; i <= z; ++i) {
            if (a[i][c]) {
                swap(a[r], a[i]);
                break;
            }
        }
        if (a[r][c] == 0) {
            ++cnt;
            continue;
        }
        for (int i = r + 1; i <= z; ++i) { // 把r后面的所有a[c]减去
            if (a[i][c]) {
                for (int j = n + 1; j >= c; --j) {
                    a[i][j] ^= a[r][j];
                }
            }
        }
        ++r;
    }
    if (r <= z) {
        for (int i = r; i <= z; ++i) {
            if (a[i][n + 1] != 0) {
                puts("0");
                return 0;
            }
        }
    }
    
    ans = 1;
    for (int i = 1; i <= cnt; ++i) { // 空行
        ans = ans * 2LL % MOD;
    }
    printf("%d\n", ans);

    return 0;
}