The king of Berland organizes a ball! nn pair are invited to the ball, they are numbered from 11 to nn. Each pair consists of one man and one woman. Each dancer (either man or woman) has a monochrome costume. The color of each costume is represented by an integer from 11 to kk, inclusive.
Let bibi be the color of the man's costume and gigi be the color of the woman's costume in the ii-th pair. You have to choose a color for each dancer's costume (i.e. values b1,b2,…,bnb1,b2,…,bn and g1,g2,…gng1,g2,…gn) in such a way that:
Let's take a look at the examples of bad and good color choosing (for n=4n=4 and k=3k=3, man is the first in a pair and woman is the second):
Bad color choosing:
Good color choosing:
You have to find any suitable color choosing or say that no suitable choosing exists.
Input
The only line of the input contains two integers nn and kk (2≤n,k≤2⋅1052≤n,k≤2⋅105) — the number of pairs and the number of colors.
Output
If it is impossible to find any suitable colors choosing, print "NO".
Otherwise print "YES" and then the colors of the costumes of pairs in the next nnlines. The ii-th line should contain two integers bibi and gigi — colors of costumes of man and woman in the ii-th pair, respectively.
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4 3
Output
YES 3 1 1 3 3 2 2 3
Input
10 4
Output
YES 2 1 1 3 4 2 3 4 4 3 3 2 2 4 4 1 1 4 3 1
Input
13 4
Output
NO
题意:题目大意,要求构造一个(ai, bi)序列,ab的值在[1, k]内,要求同一个(ai, bi)内aibi不能相同,相邻的两个ai或bi不能相同,所有的a或b不能相同,没有两个完全一样的(ai, bi)。
题解:就是个傻逼题,题目中都给你最简单的构造方式了,就是按顺序枚举两个即可,一共k*(k-1) 种
#include
using namespace std;
typedef long long ll;
ll n,m;
int main()
{
cin>>n>>m;
if(m*(m-1)=n)
{
return 0;
}
cout<=n)
{
return 0;
}
}
}
return 0;
}