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 C (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#define LIMIT_N 200000
#define LIMIT_R 1000000000

int main(int argc, char** argv) {
    int n;
    scanf("%d", &n);
    
    if (n < 2)// || n > LIMIT_N)
        return (EXIT_FAILURE);
    
    const int numRate = n;
    int Rate[numRate]; 
    int maximumDiffRate;
    int diffRate;
//    printf("about to read rates\n");
    for (int currentTime = 0; currentTime < numRate; currentTime++) {
        scanf("%d", &Rate[currentTime]);
        
        if (Rate[currentTime] < 1 || Rate[currentTime] > LIMIT_R)
//            return (EXIT_FAILURE);
            break;
        
        if (currentTime > LIMIT_N)
            break;
        
        if (currentTime > 0) { // unable to determine rate difference for only one sample.
            if (currentTime == 1) // initial declaration of rate difference
                maximumDiffRate = Rate[currentTime] - Rate[0];
            else 
                for (int someTimeBefore = 0; someTimeBefore < currentTime; someTimeBefore++) {
//                    printf("someTimeBefore and currentTime\n");
                    diffRate = Rate[currentTime] - Rate[someTimeBefore];
                    if (diffRate > maximumDiffRate)
                        maximumDiffRate = diffRate;
                }
        }
    }
    printf("%d\n", maximumDiffRate);

    return (EXIT_SUCCESS);
}