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: Runtime Error)

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

using namespace std;

#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}


typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { -1,0, 1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
//#define mp make_pair


const int MAXN = 3010;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = fact[1] = rfact[0] = rfact[1] = 1;
  for (ll i = 2; i <= n; i++) {
    fact[i] = i * fact[i - 1] % M;
    rfact[i] = mod_inverse(fact[i], M);
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
//nが大きくfactが計算できないときのほかの計算方法について書いてある
ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  ll ret = fact[n];
  ret = (ret*rfact[r]) % M;
  ret = (ret*rfact[n - r]) % M;
  return ret;
}


int n;
vi a;

ll solve1(int x) {
  assert(false);
  int now = 0;
  int cnt = 0;
  rep (i, n) {
    now ^= a[i];
    if (now == x) {
      cnt++;
      now = 0;
    }
  }

  vll dp(cnt + 1, vl(2));
  dp[0][0] = 1;
  rep (i, cnt) {
    rep (j, 2) {
      if (j == 0) {
        (dp[i + 1][0] += dp[i][j]) %= MOD;
        (dp[i + 1][1] += dp[i][j]) %= MOD;
      }
      else {
        (dp[i + 1][0] += dp[i][j]) %= MOD;
      }
    }
  }
  return dp[cnt][0];
}

ll solve21() {
  int now = 0;
  ll cnt = 0;
  rep (i, n) {
    now ^= a[i];
    if (now == 0) cnt++;
  }
  cnt--;
  return mod_pow(2, cnt);
}

ll solve22() {
  while (a.size() > 0 && a.back() == 0) a.pop_back();
  reverse(all(a));
  while (a.size() > 0 && a.back() == 0) a.pop_back();
  reverse(all(a));
  
  n = a.size();
  if (n == 0) return 0;
  
  vi b(n + 1);
  rep (i, n) {
    b[i + 1] = b[i] ^ a[i];
  }

  const int log = 20;
  
  vll dp(1 << log, vl(2));
  rep (i, 1 << log) {
    dp[i][0] = 1;
  }
  vi al(1 << log);

  int cnt = 0;
  rep1 (i, n) {
    if (b[i] == 0) {
      cnt++;
    }
    else {
      int x = b[i];
      (dp[x][0] += (cnt - al[x]) * dp[x][1] % MOD) %= MOD;
      (dp[x][1] += dp[x][0]) %= MOD;
      al[x] = cnt;
    }
  }

  ll res = 0;
  rep (i, 1 << log) {
    if (i == 0) continue;
    (res += dp[i][1]) %= MOD;
  }
  return (res + MOD) % MOD;
}

int main() {
  fio();
  cin >> n;
  a.resize(n);

  int temp = 0;
  rep (i, n) {
    cin >> a[i];
    temp ^= a[i];
  }

  if (temp != 0) {
    cout << solve1(temp) << endl;
  }
  else {
    ll res = 0;
    res += solve21();
    res += solve22();
    res %= MOD;
    cout << res << endl;
  }
}