#include <bits/stdc++.h>
#include <immintrin.h>

using namespace std;

using ll = long long;
using ld = long double;
# define x first
# define y second
# define all(x) x.begin(), x.end()
# define rall(x) x.rbegin(), x.rend()

mt19937 mt(123);

void solve();
void init();

int32_t main() {
#ifndef LOCAL
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    init();
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
}

constexpr int mod = 1'000'000'007, MAXN = 1'000'000;

int add(int a, int b) {
    return a + b >= mod ? a + b - mod : a + b;
}

int sub(int a, int b) {
    return a >= b ? a - b : a - b + mod;
}

int mul(int a, int b) {
    return 1ll * a * b % mod;
}

int bin_pow(int a, int x) {
    int res = 1;
    while (x) {
        if (x & 1) {
            res = mul(res, a);
        }
        a = mul(a, a);
        x >>= 1;
    }
    return res;
}

struct popka {
    vector<ll> basis;
    size_t sz;
    popka() {
        sz = 0;
        basis.clear();
    }
    void insert(ll x) {
        int ind = 0;
        sz++;
        for (ll i : basis) {
            int bti = 63 - __builtin_clzll(i);
            int btx = 63 - __builtin_clzll(x);
            if (bti < btx) {
                break;
            }
            ind++;
            if (bti == btx) {
                x ^= i;
            }
        }
        if (x) {
            basis.emplace(basis.begin() + ind, x);
        }
    }
    int check(ll x) {
        for (ll i : basis) {
            int bti = 63 - __builtin_clzll(i);
            int btx = 63 - __builtin_clzll(x);
            if (bti < btx) {
                break;
            }
            if (bti == btx) {
                x ^= i;
            }
        }
        if (x) {
            return 0;
        }
        return bin_pow(2, sz - basis.size());
    }
};

int inv(int x) {
	return bin_pow(x, mod - 2);
}

int f[MAXN], invf[MAXN];

void init() {
	f[0] = 1;
	for (int i = 1; i < MAXN; i++) {
		f[i] = mul(f[i - 1], i);
	}
	invf[MAXN - 1] = inv(f[MAXN - 1]);
	for (int i = MAXN - 1; i > 0; i--) {
		invf[i - 1] = mul(invf[i], i);
	}
}

int C(int n, int k) {
	return mul(f[n], mul(invf[k], invf[n - k]));
}

void solve() {
	popka p;
	ll n;
	cin >> n;
	vector<ll> arr(n);
	for (ll i = 0; i < n; i++) {
		cin >> arr[i];
	}
	ll nx = 0;
	sort(arr.begin(), arr.end());
	for (auto i : arr) {
		nx ^= i;
	}
	ll ans = 0;
	if (nx == 0) {
		ans++;
	}
	ll px = 0;
	while (!arr.empty()) {
		ll x = arr.back();
		ll c = 0;
		while (!arr.empty() && arr.back() == x) {
			c++;
			arr.pop_back();
		}
		for (ll nc = 1; nc <= c; nc++) {
			nx ^= x;
			ll now = x;
			if (nc % 2 == 1) {
				now--;
			}
			ll nans = p.check(now ^ nx);
			{
				if (px == (now ^ nx)) {
					nans = sub(nans, 1);
				}
				if (nc % 2 == 1) {
					now++;
				} else {
					now--;
				}
				if (px == (now ^ nx)) {
					nans = add(nans, 1);
				}
			}
			ans = add(ans, mul(nans, C(c, nc)));
		}
		for (ll nc = 0; nc < c; nc++) {
			px ^= x;
			p.insert(x);
		}
	}
	cout << ans << '\n';
}