Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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>

#define NUMBER_1 200000
#define NUMBER_2 199999

int main()
{
int i, j, num;
int n;
int MAX = 1-10*10*10*10*10*10*10*10*10;
int Rt[NUMBER_1];
int max[NUMBER_2];

scanf("%d", &n);

for(num=0; num<n; num++){
scanf("%d", &Rt[num]);
}

for(j=1; j<=(n-1); j++){
for(i=0; i<=(j-1); i++){
if(Rt[j]>Rt[i]){
if(Rt[j]>Rt[i]){
max[j] = Rt[j]-Rt[i];
if((max[j])>MAX){
MAX = max[j];
}
}
}
else{
max[j] = Rt[j]-Rt[i];
if((max[j])>MAX){
MAX = max[j];
}
}
}
}

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

return 0;
}