CodeForces

Triangle

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

There is a right triangle with legs of length a and b. Your task is to determine whether it is possible to locate the triangle on the plane in such a way that none of its sides is parallel to the coordinate axes. All the vertices must have integer coordinates. If there exists such a location, you have to output the appropriate coordinates of vertices.

Input

The first line contains two integers a, b (1 ≤ a, b ≤ 1000), separated by a single space.

Output

In the first line print either "YES" or "NO" (without the quotes) depending on whether the required location exists. If it does, print in the next three lines three pairs of integers — the coordinates of the triangle vertices, one pair per line. The coordinates must be integers, not exceeding 10⁹ in their absolute value.

Sample Input

Input

1 1

Output

NO

Input

5 5

Output

YES
2 1
5 5
-2 4

Input

5 10

Output

YES
-10 4
-2 -2
1 2

Source

Codeforces Round #239 (Div. 1)

首先要求在平面内找到一个直角三角形，两个边
长分别为a,b,并且三条边都不得与二维坐标轴的
平行，且点都在整数点上，首先这个三角形若是
存在的话，那么它肯定可以在这个平面上平移的
，那么我们把它给移到一二象限，然后一个点固
定在(0,0)点，这样另外两个点看一下 a,b,的范
围，那么另外两个点无非在 一二两个象限内以
(0,0)为端点的 边长为1000的正方形内，在这两
个正方形内 枚举出所有符合的点，分在两个容器
内，然后在枚举两个点 与(0,0)点是否可以组成
直角三角形，是否与坐标轴平行

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
void solve()
{int T,t,n,m,i,j,k,a,b,a1,a2=1,b1,b2;
  scanf("%d%d",&a,&b);
  for(a1=1;a1<a;a1++)
  { a2=sqrt(a*a-a1*a1);
    if(a*a==a1*a1+a2*a2)
    { b1=a1*b/a;
      b2=a2*b/a;
      if(b*b==b1*b1+b2*b2 && a2!=b1)
      { printf("YES
");
        printf("0 0
");
        printf("%d %d
",a1,a2);
        printf("%d %d
",-b2,b1);
        return;
      }
    }
  }
  printf("NO
");
}
int main()
{ solve();
}