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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>E[410], F[410];
int Deg[410], n, H, m;
void Make_Edge(int a, int b){
    E[a].push_back(b);
    F[b].push_back(a);
    Deg[b]++;
    Deg[a]--;
}
int v[410], cnt, SCC[410], ord[410];
void DFS(int a){
    int i;
    v[a]=1;
    for(i=0;i<E[a].size();i++){
        if(!v[E[a][i]])DFS(E[a][i]);
    }
    ord[++cnt] = a;
}
void DFS2(int a){
    int i;
    SCC[a] = cnt;
    for(i=0;i<F[a].size();i++){
        if(!SCC[F[a][i]])DFS2(F[a][i]);
    }
}
int main(){
    int i, a, b, c, d;
    scanf("%d%d",&n,&H);
    for(i=1;i<=n;i++){
        scanf("%d%d%d%d",&a,&b,&c,&d);
        if(!c && !d) Make_Edge(a, H+1+b);
        if(!c && d)Make_Edge(a, d);
        if(c && !d)Make_Edge(H+1+c, H+1+b);
        if(c && d)Make_Edge(H+1+c, d);
    }
    for(i=0;i<=(H+1)*2;i++){
        if((Deg[i] < 0 && i >= H+1) || (Deg[i] > 0 && i < H+1)){
            printf("NO\n");
            return 0;
        }
        while(Deg[i]!=0){
            if(Deg[i] < 0)Make_Edge(0, i);
            else Make_Edge(i, 0);
        }
    }
    m = (H+1)*2;
    for(i=0;i<=m;i++){
        if(!v[i])DFS(i);
    }
    cnt = 0;
    for(i=m+1;i>=1;i--){
        if(!SCC[ord[i]]){
            cnt++;
            DFS2(ord[i]);
        }
    }
    for(i=0;i<=m;i++){
        if(!E[i].empty() && SCC[i] != SCC[0]){
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
}