Problem:
Score : 800 points

Problem Statement

The beauty of a sequence a of length n is defined as a_1 \oplus \cdots \oplus a_n, where \oplus denotes the bitwise exclusive or (XOR).

You are given a sequence A of length N.
Snuke will insert zero or more partitions in A to divide it into some number of non-empty contiguous subsequences.

There are 2^{N-1} possible ways to insert partitions.
How many of them divide A into sequences whose beauties are all equal? Find this count modulo 10^{9}+7.

Constraints

All values in input are integers.

1 \leq N \leq 5 \times 10^5

0 \leq A_i < 2^{20}

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

3

Four ways of dividing A shown below satisfy the condition. The condition is not satisfied only if A is divided into (1),(2),(3).

(1,2,3)

(1),(2,3)

(1,2),(3)

Sample Input 2

3
1 2 2

Sample Output 2

1

Sample Input 3

32
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sample Output 3

147483634

Find the count modulo 10^{9}+7.

Sample Input 4

24
1 2 5 3 3 6 1 1 8 8 0 3 3 4 6 6 4 0 7 2 5 4 6 2

Sample Output 4

292

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
using namespace std;
///////////////////library zone!!!!!!!!!!!!!!!!!!!!!!!!!!!!
typedef long long ll;
typedef long double ld;
#define all(a) (a).begin(),(a).end()
#define EPS (1e-5)
#define bit(n,k) ((n>>k)&1)
const ll Mod = 1000000007;
const ll mod = 998244353;
struct H {
	ll x, y;
	bool operator<(const H& b) const {
		if (x != b.x) return x < b.x;
		return y < b.y;
	}
	bool operator>(const H & b) const {
		if (x != b.x) return x > b.x;
		return y > b.y;
	}
	bool operator==(const H & b) const { return x == b.x && y == b.y; }
	bool operator!=(const H & b) const { return (*this) != b; }
};
struct P {
	ll pos, cost;
	bool operator<(const P& b) const { return cost < b.cost; }
	bool operator>(const P& b) const { return cost > b.cost; }
};
struct B {
	ll to, cost;
};
struct E {
	ll from, to, cost;
	bool operator<(const E& b) const { return cost < b.cost; }
	bool operator>(const E& b) const { return cost > b.cost; }
};
template<typename T, typename U>
void chmin(T & a, U b) {
	if (a > b) a = b;
}
template<typename T, typename U>
void chmax(T & a, U b) {
	if (a < b) a = b;
}
template<typename T>
T max_0(T a) {
	if (a < 0) return 0;
	return a;
}
template<typename T>
T min_0(T a) {
	if (a > 0) return 0;
	return a;
}
ll read() {
	ll u;
	ll k = scanf("%lld", &u);
	return u;
}
ll gcd(ll i, ll j) {
	if (i > j) swap(i, j);
	if (i == 0) return j;
	return gcd(j % i, i);
}
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}//x^n%p
const ll Inf = 3023372036854775807;
const int inf = 1500000000;
#define int long long
//----------------------------------------------------
int n;
int a[600000];
int b[600000];

int u[2000000];
//0
//今までのuの総和
int v[2000000];
//今までのvの総和
int k[2000000];
signed main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		b[i + 1] = b[i] ^ a[i];
	}
	int ans = 0, sum = 0;
	if (b[n] == 0) {
		ans = 1;
		for (int i = 1; i < n; i++) {
			if (b[i] == 0) {
				ans = ans * 2 % Mod;
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		u[b[i]] = 1;
	}
	//今までにu[i]+=v[i]をした回数
	for (int i = 1; i <= n; i++) {
		if (b[i] == 0) {
			sum++;
		}
		else {
			u[b[i]] = (u[b[i]] + (v[b[i]] * (sum - k[b[i]]) % Mod)) % Mod;
			v[b[i]] += u[b[i]];
			v[b[i]] %= Mod;
			k[b[i]] = sum;
		}
	}
	if (b[n] == 0) {
		for (int i = 1; i < n; i++) {
			if (b[i] > 0) {
				ans += v[b[i]];
				ans %= Mod;
			}
		}
	}
	else {
		ans = u[b[n]];
	}
	ans += Mod;
	ans %= Mod;
	cout << ans << endl;
}