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++11 (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<vector>

class MaximumProfit{
private:
    std::vector<int> dataSet;
public:
    MaximumProfit(std::vector<int> vec);
    int getProfit();
};

MaximumProfit::MaximumProfit(std::vector<int> vec){
    this->dataSet = vec;
}

int MaximumProfit::getProfit(){
    int maxv = dataSet[1] - dataSet[0];
    for(int j = 1;j < dataSet.size();j++){
        for(int i = 0;i <= j - 1;i++){
            if(maxv < (dataSet[j] - dataSet[i])){
                maxv = dataSet[j] - dataSet[i];
            }
        }
    }
    return maxv;
}

int main(){
    
    int line;

    std::cin >> line;
    
    std::vector<int> vec;

    for(int i = 0;i < line;i++){
        int input;
        std::cin >> input;
        vec.push_back(input);
    }

    MaximumProfit mp = MaximumProfit(vec);

    std::cout << mp.getProfit() << std::endl;

    return 0;
}