Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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 <iostream>
#include <vector>

using namespace std;

typedef vector<int> VI;
#define PB push_back
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define IN(v) int v;cin>>v;
#define OUT(v) cout<<v<<endl


int main() {
  IN(N);

  VI min_v(N+1), max_v(N+1), A;
  min_v[0] = 1000000001;
  max_v[N] = 0;
  REP(i, N) {
    IN(a);
    A.PB(a);
    min_v[i+1] = min(min_v[i], a);
  }
  for (int i = N - 1; i >= 0; --i) {
    max_v[i] = max(max_v[i+1], A[i]);
  }
  int res = -1000000001;
  REP(i, N-1) {
    res = max(res, max_v[i+1] - min_v[i+1]);
  }
  OUT(res);


  return 0;
}