We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:

* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.



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Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:

* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.



Determine whether such an arrangement is possible.

Constraints

* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N


Output

If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.

Examples

Input

3 4
1 1 0 0
2 2 0 1
3 3 1 0


Output

YES


Input

4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1


Output

NO


Input

10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0


Output

YES
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;

int id[405],od[405];bool v[405];vector <int>G[405];int N,H;int a,b,c,d;

bool dfs(int tt)
{
    bool f=id[tt]!=od[tt];
    v[tt]=true;
	for(int i=0;i<G[tt].size();i++)
    {
        if(!v[G[tt][i]]) 
            f|=dfs(G[tt][i]);
    }
    return f;
}


int main()
{
	scanf("%d%d",&N,&H);
    for(int i=1;i<=N;i++) 
    {
        scanf("%d%d%d%d",&a,&b,&c,&d);
        int l,r;l= ( c==0 ? a : -c );r= ( d==0 ? -b : d );
		od[l+H]++;id[r+H]++;G[l+H].push_back(r+H);G[r+H].push_back(l+H);
    }
	bool flag=1;
	for(int i=0;flag&&i<=H;i++)if(od[i]>id[i])flag=0;
	for(int i=H+1;flag&&i<=2*H+1;i++)if(od[i]<id[i])flag=0;
	memset(v,false,sizeof(v));
    for(int i=0;flag && i<=2*H;i++)
    {if((id[i]||od[i])&&!v[i]) flag &=dfs(i); }
    if(flag)  printf("YES\n"); 
    else  printf("NO\n");
}