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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

pub mod modular {
    const M: i64 = 1000000007;

    #[derive(Debug, Clone, Copy, Default, PartialOrd, Ord, PartialEq, Eq)]
    pub struct Mod(i64);

    impl ::std::fmt::Display for Mod {
        fn fmt(&self, f: &mut ::std::fmt::Formatter) -> ::std::fmt::Result {
            write!(f, "{}", self.0)
        }
    }

    impl Mod {
        pub fn new(v: i64) -> Mod {
            Mod(v % M)
        }

        pub fn pow(self, mut r: i64) -> Mod {
            let mut k = self;
            let mut ret = 1.into();

            while r > 0 {
                if r % 2 != 0 {
                    ret = ret * k;
                }
                r /= 2;
                k = k * k;
            }

            ret
        }

        // This requires M is prime
        pub fn recip(self) -> Mod {
            self.pow(M - 2)
        }
    }

    use std::ops::*;

    impl<T: Into<Mod>> Add<T> for Mod {
        type Output = Mod;
        fn add(self, rhs: T) -> Self::Output {
            Mod::new(self.0 + rhs.into().0)
        }
    }
    impl<T: Into<Mod>> AddAssign<T> for Mod {
        fn add_assign(&mut self, rhs: T) {
            *self = *self + rhs;
        }
    }

    impl<T: Into<Mod>> Sub<T> for Mod {
        type Output = Mod;
        fn sub(self, rhs: T) -> Self::Output {
            Mod::new(self.0 - rhs.into().0 + M)
        }
    }
    impl<T: Into<Mod>> SubAssign<T> for Mod {
        fn sub_assign(&mut self, rhs: T) {
            *self = *self - rhs;
        }
    }

    impl<T: Into<Mod>> Mul<T> for Mod {
        type Output = Mod;
        fn mul(self, rhs: T) -> Self::Output {
            Mod::new(self.0 * rhs.into().0)
        }
    }
    impl<T: Into<Mod>> MulAssign<T> for Mod {
        fn mul_assign(&mut self, rhs: T) {
            *self = *self * rhs;
        }
    }

    impl<T: Into<Mod>> Div<T> for Mod {
        type Output = Mod;
        fn div(self, rhs: T) -> Self::Output {
            self * rhs.into().recip()
        }
    }
    impl<T: Into<Mod>> DivAssign<T> for Mod {
        fn div_assign(&mut self, rhs: T) {
            *self = *self / rhs;
        }
    }

    impl Neg for Mod {
        type Output = Mod;
        fn neg(self) -> Self::Output {
            Mod(0) - self
        }
    }

    impl<T: ::std::convert::Into<i64>> ::std::convert::From<T> for Mod {
        fn from(v: T) -> Self {
            Mod::new(v.into())
        }
    }
}

use modular::Mod;
use std::collections::BTreeMap;

fn main() {
    input! {
        n: usize,
        a: [i32; n],
    }

    let mut suml = vec![0; n + 1]; // sum [0..i)
    let mut sumr = vec![0; n + 1]; // sum [i..n)

    for i in 0..n {
        suml[i + 1] = suml[i] ^ a[i];
    }

    for i in (0..n).rev() {
        sumr[i] = sumr[i + 1] ^ a[i];
    }

    // dbg!(&suml);

    let mut mm = BTreeMap::new();
    let ans = solve(0, -1, n, &suml, &sumr, &mut mm);

    println!("{}", ans);
}

fn solve(
    i: usize,
    tgt: i32,
    n: usize,
    suml: &[i32],
    sumr: &[i32],
    tbl: &mut BTreeMap<(usize, i32), Mod>,
) -> Mod {
    if i >= n {
        return 1.into();
    }

    let key = (i, tgt);
    if let Some(ret) = tbl.get(&key) {
        return *ret;
    }

    let mut ret = 0.into();

    for j in i + 1..n + 1 {
        let x = suml[i] ^ suml[j];
        let y = sumr[j];

        // dbg!((i, j, x));

        if (tgt == -1 || x == tgt) && (y == x || y == 0) {
            ret += solve(j, x, n, suml, sumr, tbl);
        }
    }

    tbl.insert(key, ret);
    ret
}