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 <bits/stdc++.h> //全てのヘッダファイルをインクルード

//ループ
#define REP(i, n) for(int i = 0; i < n; i++) //普通のループ
#define REPR(i, n) for(int i = n; i >= 0; i--) //逆ループ
#define FOR(i, m, n) for(int i = m; i < n; i++) //最初の数値を指定したループ

//型名省略
typedef long long ll;

//配列要素数
static const int MAX = 200000;

using namespace std;

//#include "./lib/string.h"

int main(){
    ll n, R[MAX];

    cin >> n;
    REP(i, n) cin >> R[i];

    ll maxv = -2000000000;
    ll minv = R[0];

    FOR(i, 1, n){
        maxv = max(maxv, R[i]-minv);
        minv = min(minv, R[i]);
    }

    cout << maxv << endl;

    return 0;

    return 0;
}