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
//AtCoder Grand Contest 017 E - Jigsaw

#include <cstdio>
#include <iostream>
#include <cmath>

using namespace std;

int qr(){int ret = 0,f=1;char ch =getchar();while(!isdigit(ch))f=ch=='-'?-1:1,ch=getchar();while(isdigit(ch))ret=ret*10+ch-'0',ch=getchar();return ret*f;}

#define maxN 100005
#define maxH 205

int indegree[maxH<<2],outdegree[maxH<<2];
int active[maxH<<2];
void read(int x,int y){
    indegree[y]++;
    outdegree[x]++;
}
int f[maxH<<1],pd[maxH<<1];
int gf(int x){
    if(f[x] == x) return x;
    else return f[x] = gf(f[x]);
}
int main(){
    int N = qr(),H = qr();
    for(int i = 0;i<=(H<<1);i++)
        f[i] = i;
    int t1,t2,t3,t4,t5,t6;//上 t + maxN 下 t + maxN + maxH
    for(int i = 1;i<=N;i++){
        t1 = qr(),t2 = qr(),t3 = qr(),t4 = qr();
        if(!t3) t5 = t1;else t5 = -t3;
        if(!t4) t6 = -t2;else t6 = t4;
        t5 += H,t6 += H;
        active[t5] = active[t6] = 1;
        indegree[t6]++;outdegree[t5]++;
        f[gf(t5)] = gf(t6);
    }
    int mark=1;
    for(int i = 0;i<=H;i++){
        if(indegree[i]<outdegree[i])
            mark=0;
    }
    for(int i = H+1;i<=(H<<1);++i){
        if(indegree[i]>outdegree[i])
            mark=0;
    }
    for(int i = 0;i<=(H<<1);i++){
        pd[gf(i)] |= (indegree[i] != outdegree[i]);
    }
    pd[H] = 1;
    for(int i = 1;i<=(H<<1);i++){
        if(active[i] && !pd[gf(i)])
            mark=0;
    }
    if(mark)
        cout << "YES";
    else cout << "NO";
    return 0;
}