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.



<image>

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<bits/stdc++.h>
#define pb push_back
using namespace std;
const int N=1e5+10;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int n,h,in[500],out[500];
vector<int> E[500];
bool vis[500],fl=0;
void dfs(int u) {
    vis[u]=1;
    if(in[u]!=out[u]) fl=1;
    for(int v:E[u])
        if(!vis[v]) dfs(v);
}
int main() {
    n=gi(),h=gi();
    for(int i=1;i<=n;i++) {
        int a=gi(),b=gi(),c=gi(),d=gi(),l,r;
        if(c==0) l=a; else l=-c;
        if(d==0) r=-b; else r=d;
        l+=h,r+=h;
        E[l].pb(r),E[r].pb(l),++out[l],++in[r];
    }
    for(int i=0;i<=2*h;i++) {
        if(i<h&&in[i]<out[i]) return puts("NO"),0;
        if(i>h&&in[i]>out[i]) return puts("NO"),0;
    }
    for(int i=0;i<=2*h;i++)
        if(i!=h&&!vis[i]&&in[i]+out[i]) {
            fl=0,dfs(i);
            if(!fl) return puts("NO"),0;
        }
    puts("YES");
    return 0;
}
