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)

// Maximum Profit
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<stdbool.h>

const int MAX_N = 200000;

int main(void)
{
    int tmp;
    int max_val = 0;
    int input_num, ary_price[MAX_N];

    for(int i = 0; i < MAX_N; i++)
    {
        ary_price[i] = 0;
    }

    scanf("%d", &input_num);

    for(int i = 0; i < input_num; i++)
    {
        scanf("%d", &ary_price[i]);
    }

    max_val = ary_price[input_num - 1] - ary_price[input_num - 2];

    for (int i = input_num - 1; i > 0; i--)
    {
        for(int j = i - 1; j >= 0; j--)
        {
            tmp = ary_price[i] - ary_price[j];

            if( max_val < tmp )
            {
                max_val = tmp;
            }
        }
    }

    printf("%d\n", max_val);

    return 0;
}