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>
#define MAX 200000
int main()
{
  int i,j,maxv,minv,R[MAX],n;

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

maxv=R[1]-R[0];

for(j=1;j<=n-1;j++){
  for(i=0;i<=j-1;i++){
    if(maxv>R[j]-R[i]){
      maxv=maxv;
    }
    if(maxv<R[j]-R[i]){
      maxv=R[j]-R[i];
    }
  }
 }

  minv=R[0];
  for(j=1;j<=n-1;j++){
    if(maxv>R[j]-minv){
      maxv=maxv;
    }
    if(maxv<R[j]-minv){
      maxv=R[j]-minv;
    }
    if(minv>R[j]){
      minv=R[j];
    }
    if(minv<R[j]){
      minv=minv;
    }
  }
  printf("%d\n",maxv);

  return 0;
}