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)

#if 0
#include <stdio.h>
//复杂度为n^2的算法
int main()
{
    int n;
    int A[2000010];
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&A[i]);
    }
    int max=A[1]-A[0];
    for(int i=0;i<n;i++)
    {
        int j=i+1;
        for(;j<n;j++)
        {
           if(A[j]-A[i]>max)max=A[j]-A[i];
        }
    }
    printf("%d\n",max);
    return 0;
}
#elif 0
#include<stdio.h>
int main()
{
    int i,n,a[200010],ans,min;
    scanf("%d",&n);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    min=a[0];
    ans=a[1]-a[0];
    for(i=1;i<n;i++)
    {
        if(a[i]<min) min=a[i];
        else if(a[i]-min>ans) ans=a[i]-min;
    }
    printf("%d\n",ans);
    return 0;
}

#elif 0
#include<stdio.h>
int main()
{
    int n,min,max,m,m2;
    scanf("%d%d%d",&n,&m,&m2);
    min=m;
    max=m2-min;
    for(int i=2;i<n;i++)
    {
        scanf("%d",&m);
        if(m-min>max)max=m-min;
        if(m<min)min=m;
        printf("min=%d max=%d\n",min,max);
    }
    printf("%d\n",max);
   return 0;
}
#elif 1

#include <stdio.h>
int main()
{
    int A[200010],n,min,max;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&A[i]);
    }
    min=A[0];
    max=A[1]-min;
    for(int i=1;i<n;i++)
    {
        if(A[i]-min>max)max=A[i]-min;
        if(A[i]<min)min=A[i];
    }
    printf("%d\n",max);
    return 0;
}

#endif
/*
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*/