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>

int main(){

    int t; //first line

    scanf("%d",&t);

    int r[t];
    int i = 0;
    int p;
    int j;


    while(i < t){
        scanf("%d",&r[i]);
        i++;
    }

    // p = r[1] - r[0];
    // for(i=0; i<t; i++){
    //     for(j=i+1; j<t; j++){
    //         if(p < (r[j] - r[i])){
    //             p = r[j] - r[i];
    //         }
    //     }
    // }

    i = 0;
    p = r[1] -r[0];
    for(j=i+1; j<t; j++){
        if(p < (r[j] - r[i])){
            p = r[j] - r[i];
        }
//        printf("%d\n",p);
    }
    

    for(i=1; i<t; i++){
        if(r[i] < r[i-1]){
            for(j=i+1; j<t; j++){
                if(p < (r[j] - r[i])){
                    p = r[j] - r[i];
                }
            }
        }
    }




    printf("%d\n",p);

    return 0;

}