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)

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#pragma warning(disable:4786)
#pragma warning(disable:4996)
#include <ctime>
#include<list>
#include <numeric>
#include<bitset>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<functional>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<utility>
#include<fstream>
#include<sstream>
#include<cmath>
#include<stack>
#include<assert.h>
#include<unordered_map>
#include<unordered_set>
using namespace std;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
	~debug() { cerr << endl; }
	template<class c> typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
		cerr << boolalpha << i; return *this;
	}

	template<class c, int=0> typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
		return *this << range(begin(i), end(i));
	}

sim, class b dor(pair < b, c > d) {
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
#else
	sim dor(const c&) { ris; }
#endif
};
#define watch(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define MEM(a, b) memset(a, (b), sizeof(a))
#define CLR(a) memset(a, 0, sizeof(a))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define ABS(X) ( (X) > 0 ? (X) : ( -(X) ) )
#define S(X) ( (X) * (X) )
#define SZ(V) (int )V.size()
#define FORN(i, n) for(int i = 0; i < n; i++)
#define FORAB(i, a, b) for(int i = a; i <= b; i++)
#define ALL(V) V.begin(), V.end()
#define IN(A, B, C)  ((B) <= (A) && (A) <= (C))
#define AIN(A, B, C) assert(IN(A, B, C))

//typedef int LL;
typedef long long int LL;
//typedef __int128 LLL;
typedef long long LLL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PII > VP;
typedef vector<PLL > VPL;
typedef vector<double> VD;
typedef long double ld;

int a[500005];
int dp[2][1 << 20];
const LL MOD = 1000000007;
int lz[1 << 20];

void solve(int ks) {
	int n, now = 0;
	int cnt = 0;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		now ^= a[i];
		if (now == 0) cnt++;
	}

	LL ans = 0;
	if (now == 0) {
		ans = 1;
		for (int i = 1; i < cnt; i++) {
			ans = (ans * 2) % MOD;
		}
	}

	for (int i = 1; i < (1 << 20); i++) lz[i] = 0, dp[1][i]=1;
	int cur = 0;

	now = 0;
	for (int i = 1; i <= n; i++) {
		now ^= a[i];
		if (now) {
			int last = lz[now];
			dp[1][now] += (cur - last) * dp[0][now];
			dp[1][now] %= MOD;
			dp[0][now] += dp[1][now];
			dp[0][now] %= MOD;
			lz[now] = cur;
			if (i == n) {
				ans += dp[1][now];
				ans %= MOD;
			}
		}
		else {
			cur++;
			if (i == n) {
				for (int j = 1; j < (1<<(20)); j++) {
					ans += dp[0][j];
					ans %= MOD;
				}
			}
		}
	}
	printf("%lld\n", ans);
}

int main()
{
#ifdef LOCAL
	double start_time = clock();
	freopen("C:\\Home\\ContestCodes\\sample.in", "r", stdin);
	//	freopen("out.out", "w", stdout);
#endif


	if (0) {
		int T;
		scanf("%d", &T);
		for (int ks = 1; ks <= T; ks++) solve(ks);
	}
	else {
		solve(0);
	}

#ifdef LOCAL
	double end_time = clock();
	fprintf(stderr, "Time = %lf\n", (end_time - start_time) / CLOCKS_PER_SEC);
#endif

	return 0;
}