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>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

const int N = 4e5 + 10, H = 505;
int n, h, deg[N];
bool tag, vis[N], mark[N];
vector<int> nxt[N];

void add(int x, int y) {
    x += H, y += H;
    mark[x] = mark[y] = 1;
    nxt[x].push_back(y);
    nxt[y].push_back(x);
    ++deg[y], --deg[x];
}

bool dfs(int x) {
    if ((x >= H && deg[x] > 0) || (x < H && deg[x] < 0)) return 0;
    vis[x] = 1;
    if (deg[x]) tag = 1;
    for (int i = 0; i < nxt[x].size(); i++) {
        int y = nxt[x][i];
        if (!vis[y]) if (!dfs(y)) return 0;
    }
    return 1;
}

int main() {
    cin >> n >> h;
    rep(i, 1, n) {
        int a, b, c, d; scanf("%d%d%d%d", &a, &b, &c, &d);
        add(c ? -c : a, d ? d : -b);
    }
    rep(i, 1, H << 1) if (mark[i] && !vis[i]) {
        tag = 0;
        if (!dfs(i) || !tag) return puts("NO"), 0;
    }
    puts("YES");
    return 0;
}
