Problem:
Score : 400 points

Problem Statement

There are N mountains in a circle, called Mountain 1, Mountain 2, ..., Mountain N in clockwise order. N is an odd number.

Between these mountains, there are N dams, called Dam 1, Dam 2, ..., Dam N. Dam i (1 \leq i \leq N) is located between Mountain i and i+1 (Mountain N+1 is Mountain 1).

When Mountain i (1 \leq i \leq N) receives 2x liters of rain, Dam i-1 and Dam i each accumulates x liters of water (Dam 0 is Dam N).

One day, each of the mountains received a non-negative even number of liters of rain.

As a result, Dam i (1 \leq i \leq N) accumulated a total of A_i liters of water.

Find the amount of rain each of the mountains received. We can prove that the solution is unique under the constraints of this problem.

Constraints

All values in input are integers.

3 \leq N \leq 10^5-1

N is an odd number.

0 \leq A_i \leq 10^9

The situation represented by input can occur when each of the mountains receives a non-negative even number of liters of rain.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N integers representing the number of liters of rain Mountain 1, Mountain 2, ..., Mountain N received, in this order.

Sample Input 1

3
2 2 4

Sample Output 1

4 0 4

If we assume Mountain 1, 2, and 3 received 4, 0, and 4 liters of rain, respectively, it is consistent with this input, as follows:

Dam 1 should have accumulated \frac{4}{2} + \frac{0}{2} = 2 liters of water.

Dam 2 should have accumulated \frac{0}{2} + \frac{4}{2} = 2 liters of water.

Dam 3 should have accumulated \frac{4}{2} + \frac{4}{2} = 4 liters of water.

Sample Input 2

5
3 8 7 5 5

Sample Output 2

2 4 12 2 8

Sample Input 3

3
1000000000 1000000000 0

Sample Output 3

0 2000000000 0

Solution in Rust (1.15.1) (judge verdict: Accepted)

mod stdin_reader {
  use std::io::stdin;
  use std::io::Read;

  pub struct StdinReader {
    pub buf: Vec<String>,
    pub cursor: usize,
  }

  pub fn new() -> StdinReader {
    StdinReader::new()
  }

  impl StdinReader {
    pub fn new() -> Self {
      let mut s = String::new();
      let stdin = stdin();
      stdin.lock().read_to_string(&mut s).unwrap();
      let buf = s.split_whitespace().map(|s| s.to_string()).collect();
      return StdinReader { buf: buf, cursor: 0 }
    }

    pub fn read_all(&self) -> Vec<String> {
      return self.buf.clone()
    }

    pub fn next(&mut self) -> Option<String> {
      let val = self.buf.get(self.cursor).cloned();
      self.cursor += 1;
      return val
    }

    pub fn as_i64(&mut self) -> Option<i64> {
      self.next().and_then(|s| s.parse::<i64>().ok())
    }

    pub fn as_f64(&mut self) -> Option<f64> {
      self.next().and_then(|s| s.parse::<f64>().ok())
    }

    pub fn as_usize(&mut self) -> Option<usize> {
      self.next().and_then(|s| s.parse::<usize>().ok())
    }
  }
}

mod utils {
  pub fn is_int(n: f64) -> bool {
    let trunc = (n.trunc() as i64) as f64;
    return (n - trunc) < 0.0001
  }
}

fn main() {
  // stdout
  use std::io::{stdout, Write, BufWriter};
  let stdout = stdout();
  let mut writer = BufWriter::new(stdout.lock());

  // stdin
  let mut reader = stdin_reader::new();

  let n = reader.as_usize().unwrap();
  let a_list: Vec<i64> = (0..n).map(|_| reader.as_i64().unwrap()).collect();
  let sum: i64 = a_list.iter().sum();
  let even_a_sum: i64 = a_list.iter().enumerate().filter(|&(i, val)| (i+1) % 2 == 0).map(|(i, val)| val).sum();
  let mut x: i64 = sum - 2 * even_a_sum;
  let mut result = format!("{}", x);

  for i in 0..(n-1) {
    x = 2 * a_list[i] - x;
    result.push_str(&format!(" {}", x));
  }

  writeln!(writer, "{}", result);
}