Codeforces Round #353 (Div. 2) B. Restoring Painting

Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.

• The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
• The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
• Four elements a, b, c and d are known and are located as shown on the picture below.

Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.

Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.

Input

The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.

Output

Print one integer — the number of distinct valid squares.

Examples

Input

2 1 1 1 2

Output

2

Input

3 3 1 2 3

Output

6

分析：可以无视掉中间的，求出左上角的取值范围（l,r）然后乘上n就行，没有判断r - l + 1>0,rating成功达到了人生新低。。。

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
long long n,a,b,c,d;
int main()
{
	cin.sync_with_stdio(false);	
	cin>>n>>a>>b>>c>>d;
	long long A[3] = {b-c,a-d,a-d+b-c};
	sort(A,A+3);
	long long l = max(1ll,1ll - A[0]);
	long long r = min(n,n - A[2]);
	if(l > n || r <= 0 || r - l + 1 < 0) cout<<0<<endl;
	else cout<<(r-l+1ll)*n;
}