Problem:
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Maximum Profit

You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen.

Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .

Input

The first line contains an integer $n$. In the following $n$ lines, $R_t$ ($t = 0, 1, 2, ... n-1$) are given in order.

Output

Print the maximum value in a line.

Constraints

$2 \leq n \leq 200,000$

$1 \leq R_t \leq 10^9$

Sample Input 1

6
5
3
1
3
4
3

Sample Output 1

3

Sample Input 2

3
4
3
2

Sample Output 2

-1

Solution in Rust (judge verdict: Accepted)

use std::io::BufRead;

fn main() {
    let stdin = std::io::stdin();
    
    let mut buf: String = String::new();

    stdin.lock().read_line(&mut buf).expect("read_line()");
    stdin.lock().read_line(&mut buf).expect("read_line()");
    let vec: Vec<&str> = buf.split_whitespace().collect();

    // The number of elements (r0, r1, ..., r_n-1)
    let n = vec[0].parse().unwrap();

    // The old value r_t-1 at r_t
    let mut r_old: i32 = vec[1].parse().unwrap(); // r_t-1

    // Minimum value of range r0, r1, ..., rt
    let mut min = r_old;

    // Current maximum gain
    let mut gain_max: i32 = -1000000000;

    // Current gain
    let mut gain = 0;

    let mut buf = String::new();
    for _t in 1..n {
        stdin.lock().read_line(&mut buf).expect("read_line()");
    }
    let r_series: Vec<&str> = buf.split_whitespace().collect();
    for t in 1..n {
        let rt = r_series[t-1].parse().unwrap();
        let diff = rt - r_old;
        if min > rt {
            min = rt;
            if gain_max < diff {
                gain_max = diff;
            }
            gain = 0;
        } else {
            gain += diff;
            if gain_max < gain {
                gain_max = gain;
            }
        }
        r_old = rt;
    }
    println!("{}", gain_max);
}