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;

const int Nmax = 500;

int n, H, i, x, y, A, B, C, D, grad[Nmax];
bool ok, used[Nmax];
vector<int> v[Nmax];

void dfs(int node)
{
    used[node] = 1;
    ok |= (grad[node] != 0);
    for(auto it : v[node])
        if(!used[it]) dfs(it);
}

int main()
{
   // freopen("input", "r", stdin);

    cin >> n >> H;
    for(i=1; i<=n; ++i)
    {
        cin >> A >> B >> C >> D;
        x = (C ? C : A + H);
        y = (D ? D + H : B);

        v[x].push_back(y);
        v[y].push_back(x);
        ++grad[x]; --grad[y];
    }

    ok = 1;
    for(i=1; i<=H && ok; ++i) ok &= (grad[i] <= 0);
    for(; i<= 2*H && ok; ++i) ok &= (grad[i] >= 0);

    for(i=1; i <= 2 * H && ok; ++i)
        if(!used[i] && v[i].size())
        {
            ok = 0;
            dfs(i);
        }

    cout << (ok ? "YES" : "NO") << '\n';

    return 0;
}
