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: Accepted)


#include<stdio.h>

#define INPUT_MAX_NUM (200000)

int main(void)
{
  //input num
  int num = 0;
  //  printf("input num\n");
  scanf("%d", &num);

  //input value
  int r[INPUT_MAX_NUM];
  for (int i = 0; i < num; i++ ) {
    scanf("%d", &r[i]);
  }

  //max value of diff
  /*
  int max_diff;
  max_diff = r[1] - r[0];
  for (int i = 1; i < num; i++ ) {
    for (int j = i; j < num; j++ ) {
      if ( max_diff < r[j] - r[i-1] ) {
        max_diff = r[j] - r[i-1];
      }
    }
  }
  */

  int max_diff = r[1] - r[0];
  int min = r[0];
  for (int i = 1; i < num; i++ ){
    if ( max_diff < r[i] - min ) {
      max_diff = r[i] - min;
    }
    if ( min > r[i] ) {
      min = r[i];
    }
  }
  //printf("max_diff = %d\n", max_diff);
  printf("%d\n", max_diff);

  return 0;
}