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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int MX=411, inf=2e9;

int n, h;

int indeg[MX], outdeg[MX];

int U[MX];
int find(int x){
    return x==U[x] ? x : U[x]=find(U[x]);
}
void unite(int x, int y){
    U[find(y)]=find(x);
}

bool solve(){
    bool st[MX]={}, ed[MX]={};
    for(int i=1; i<=h; i++){
        if(indeg[i]<outdeg[i]) return false;
        if(indeg[i]==outdeg[i]) continue;
        ed[find(i)]=true;
    }
    for(int i=h+1; i<=2*h; i++){
        if(indeg[i]>outdeg[i]) return false;
        if(indeg[i]==outdeg[i]) continue;
        st[find(i)]=true;
    }
    for(int i=1; i<=2*h; i++){
        if(indeg[i]==0 && outdeg[i]==0) continue;
        int x=find(i);
        if(!st[x] || !ed[x]) return false;
    }
    return true;
}

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    cin>>n>>h; h++;
    iota(U, U+2*h+1, 0);
    for(int i=1; i<=n; i++){
        int a, b, c, d; cin>>a>>b>>c>>d;
        int x = (c==0 ? h+a : c), y = (d==0 ? b : h+d);
        unite(x, y);
        outdeg[x]++, indeg[y]++;
    }
    cout<<(solve() ? "YES" : "NO");
    return 0;
}