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>
int main(){
  int a,i,j;
  int max,min;
  int *num;
  for(;;){
  scanf("%d",&a);
  if(a>=2 && a<=200000)
    break;
  }
  num=(int *)malloc(a * sizeof(int));
  for(i=0;i<a;i++){
    for(;;){
    scanf("%d",&num[i]);
    if(num[i]>=1 && num[i]<=1000000000)
      break;
    }
  }
  max=num[1]-num[0];
  for(j=1;j<a;j++){
    for(i=0;i<j;i++){
      if(max>(num[j]-num[i]))max=max;
      else max=num[j]-num[i];
    }
  }
  min=num[0];
  for(j=1;j<a;j++){
    if(max>(num[j]-min))max=max;
    else if(max<(num[j]-min))max=num[j]-min;
    if(min<num[j])min=min;
    else if(min>num[j])min=num[j];
    }
  printf("%d\n",max);
  free(num);
  return 0;
}