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 JAVA (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
import static java.lang.Math.*;

public class Main {
  public static void main(String[] args) {
    int i, j;
    Scanner sc = new Scanner(System.in); 
    int n = Integer.parseInt(sc.next());;
    int r[] = new int[n];
    for (i=0; i<n; i++)
      r[i] = Integer.parseInt(sc.next());;

    //res???max-min???????????????(???????????????mini?????´??°??§???????????¨???????????????)
    int res = r[1] - r[0];
    int mini = r[0];
    for (j=1; j<n; j++) {
      if (res < r[j] - mini) res = r[j] - mini;
      if (r[j] < mini) mini = r[j];
    }

    System.out.println(res);
  }
}

/*
  //O(n^2):TLE
  int res = r[1] - r[0];
  for (i=0; i<n-1; i++)
  for (j=i+1; j<n; j++)
    res = (res < r[j] - r[i]) ? r[j] - r[i] : res; 
  printf("%d\n", res);
*/
   
/*
  //WA(mini???????¨??????´??°?????????maxj-mini????????????????????¨?????????)
  int maxj = r[1];
  int mini = r[0];
  for (j=1; j<n; j++) {
    if (maxj-mini < r[j]-mini) {
    maxj = r[j];
    if (j < n-1 && r[j] < mini) mini = r[j];
    }
  //    if (j < n-1 && r[j] < mini) mini = r[j];
  }
  printf("%d\n", maxj-mini);
*/