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;

int N,H;
vector<int> G[410],R[410];
bool vis[410];
vector<int> ele;

void DFS(int v)
{
    if(vis[v])return;
    vis[v] = true;
    ele.push_back(v);
    for(int i = 0; i < G[v].size(); i++)DFS(G[v][i]);
    for(int i = 0; i < R[v].size(); i++)DFS(R[v][i]);
    return;
}

int main()
{
    scanf("%d%d",&N,&H);
    for(int i = 0; i < N; i++)
    {
        int a,b,c,d;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        int s,t;
        if(c == 0)s = a;
        else s = c + 200;
        if(d == 0)t = b + 200;
        else t = d;
        G[s].push_back(t);
        R[t].push_back(s);
    }

    for(int i = 0; i < 410; i++)if(!vis[i])
    {
        if(G[i].size() == 0 && R[i].size() == 0)continue;
        ele.clear();
        DFS(i);
        bool flag = true;
        for(int j = 0; j < ele.size(); j++)
        {
            int now = ele[j];
            if(G[now].size() == R[now].size())continue;
            flag = false;
            if(G[now].size() > R[now].size() && now >= 201)
            {
                printf("NO\n");
                return 0;
            }
            if(G[now].size() < R[now].size() && now < 201)
            {
                printf("NO\n");
                return 0;
            }
        }
        if(flag)
        {
            printf("NO\n");
            return 0;
        }
    }

    printf("YES\n");
    return 0;
}
