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>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int a[200005],b[200005],c[200005];
int main(){
    int t;
    while( scanf("%d",&t)!=EOF){
        int i;
        for(i=1;i<=t;i++){
            scanf("%d",&a[i]);
        }
        b[0]=1e9+1;
        for(i=1;i<=t;i++){
            if(a[i]<b[i-1]) b[i]=a[i];
            else b[i]=b[i-1];
        }
        c[t+1]=0;
        for(i=t;i>=1;i--){
            if(a[i]>c[i+1]) c[i]=a[i];
            else c[i]=c[i+1];
        }
        int max_num=-1e9-1;
        for(i=1;i<t;i++){
            if(c[i+1]-b[i]>max_num) max_num=c[i+1]-b[i];
        }
        printf("%d\n",max_num);
    }
    return 0;
}