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++14 (judge verdict: Accepted)

//include
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <iomanip>
#include <map>
#include <sstream>
#include <utility>
//マクロとか
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i < (int)(n) + 1; i++)
#define all(vecto) vecto.begin(), vecto.end()
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
using ll = long long;

//自前の関数たち
//最大公約数
ll GCD(ll a, ll b)
{
    if (b == 0)
        return a;
    else
        return GCD(b, a % b);
}

//最小公倍数
ll LCM(ll a, ll b)
{
    return (a * b) / GCD(a, b);
}

//namespace my
namespace my {
    //配列の最大値or最小値を数値で返す関数
    int max_element(std::vector<int> vec){
        std::sort(vec.begin(), vec.end());
        return vec.at(vec.size() - 1);
    }
    int min_element(std::vector<int> vec){
        std::sort(vec.begin(), vec.end());
        return vec.at(0);
    }
}

//yes,noを出力するやつ
void ans(bool qus)
{
    if (qus == true)
        std::cout << "Yes" << std::endl;
    else
        std::cout << "No" << std::endl;
}

void Ans(bool qust)
{
    if (qust == true)
        std::cout << "YES" << std::endl;
    else
        std::cout << "NO" << std::endl;
}

//std名前空間を使用
using namespace std;

////////////////////////////////////////////////////////////////////////////////
//スタート


int main()
{
    int n;
    cin >> n;
    vector<int> vec(n);
    rep(i,n){
        cin >> vec[i];
    }

    int maxv = -2000000000;
    int minv = vec[0];

    for (int i = 1; i < n; i++)
    {
        maxv = max(maxv, vec[i]-minv);
        minv = min(minv, vec[i]);
    }

    cout << maxv << endl;
}