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 <stdio.h>

//問題文より、入力されるデータの最大個数
#define MAXCOUNT 200000

int main(void)
{
	/*
	nは入力されるデータの個数を格納
	new_price - base_priceで利益を求め、diff_priceに代入する
	*/
	int i, n, base_price, new_price, diff_price;
	int price[MAXCOUNT];

	//データの個数を入力
	scanf("%d", &n);

	//データを入力
	for(i = 0; i < n; i++){
		scanf("%d", &price[i]);
	}

	//最大の利益を求める
	for(i = 0; i < (n - 1); i++){
		//ループの最初ならば初期値を代入
		if(i == 0){
			base_price = price[i];
			new_price = price[i + 1];
			diff_price = new_price - base_price;
		}
		//基準価格は最小にする
		else if(base_price > price[i]){
			base_price = price[i];
			new_price = price[i + 1];
			//最大利益が出るならば、更新する
			if(diff_price < (new_price - base_price)){
				diff_price = new_price - base_price;
			}
		}
		//変動後の価格は最大にする
		else if(new_price < price[i + 1]){
			new_price = price[i + 1];
			//最大利益が出るならば、更新する
			if(diff_price < (new_price - base_price)){
				diff_price = new_price - base_price;
			}
		}
	}

	//結果を出力
	printf("%d\n", diff_price);
	
	return 0;
}