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: Accepted)

#include <stdio.h>
#define N 200000

int main(){
  int n, num[N]={0};
  int i = 0, j, max, minnum;
  
  scanf("%d", &n);
  
  while(i<n){
    scanf("%d", &num[i]);
    i++;
  }
  
  max = num[1] - num[0];
  minnum= num[0];
    
  for(i = 0; i < n; i++){
    if(minnum < num[i]){
      continue;
    }

    minnum = num[i];
    
    for(j = i+1; j < n; j++){
      if(max < num[j] - minnum){
	max = num[j] - minnum;
      }
    }  
  }
  
  /*  
  max = num[1];
  min = num[0];
  for(i = 1; i < n; i++){
    if(max <= num[i]){
      max = num[i];
      
      for(j = 0; j < i; j++){
	if(min > num[j]){
	  min = num[j];
	}
	
      }
    }
  }
  */
  
  printf("%d\n", max);
  
  return 0;
}