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 <string>
# include <algorithm>
# include <vector>
# include <cstring>
# include <stdio.h>
# include <map>
# include <queue>

# define ll long long
# define pii pair<int,int>
# define FOR(a,b) for(int a=1; a<=b; a++)
# define REP(a,b) for(int a=0; a<b; a++)
# define FORU(a,b,c) for(int a=b; a<=c; a++)
# define FORD(a,b,c) for(int a=b; a>=c; a--)

using namespace std;

bool flag[505];
int N, H, par[505], indeg[505], outdeg[505], ccsize[505];

int findpar(int x){
    return x == par[x] ? x : par[x] = findpar(par[x]);
}

void merge(int u,int v){
    u = findpar(u);
    v = findpar(v);

    ccsize[v]++;
    if(u != v){
        ccsize[v] += ccsize[u];
        par[u] = v;
    }
}

int main(){

    ios :: sync_with_stdio(false);
    cin >> N >> H;
    REP(i,2*H+1) par[i] = i;

    FOR(i,N){
        int a,b,c,d;
        cin >> a >> b >> c >> d;
        int u = c ? -c : a;
        int v = d ? d : -b;

        u += H;
        v += H;

        outdeg[u]++;
        indeg[v]++;
        merge(u, v);
    }

    bool ok = 1;
    REP(i,2*H+1){
        if(i<H && indeg[i] < outdeg[i]) ok = 0;
        if(i>H && indeg[i] > outdeg[i]) ok = 0;
        if(indeg[i] != outdeg[i]) flag[findpar(i)] = 1;
    }

    REP(i,2*H+1){
        if(ccsize[i]>0 && !flag[findpar(i)]) ok = 0;
    }

    cout << (ok ? "YES" : "NO") << endl;

    return 0;
}