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 <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_map>
#include <functional>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cstdlib>

using namespace std;
typedef long long llong;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<llong, llong> pll;

void fail(int i) {
    printf("NO\n", i);
    exit(0);
}

int n, h;
struct block {
    int l, r;
    void scan() {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        if (c == 0) l = a << 1;
        else l = c << 1 | 1;
        if (d == 0) r = b << 1 | 1;
        else r = d << 1;
    }
} bs[100000];

int par[402];
int find(int x) {
    if (par[x] != x) return par[x] = find(par[x]);
    return x;
}

int cnt[402];
int ex[402];
int zr[402];
int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	for (int i = 0; i < 402; ++i) par[i] = i;
    cin >> n >> h;
    for (int i = 0; i < n; ++i) {
        bs[i].scan();
        ++cnt[bs[i].l];
        --cnt[bs[i].r];
        ex[bs[i].l] = 1;
        ex[bs[i].r] = 1;
        par[find(bs[i].l)] = find(bs[i].r);
    }
    for (int i = 0; i < 402; ++i) {
        if (cnt[i] < 0 && ((i ^ 1) & 1)) fail(0);
        if (cnt[i] > 0 && ((i ^ 0) & 1)) fail(1);
        if (cnt[i] != 0) zr[find(i)] = 1;
    }
    for (int i = 0; i < 402; ++i) {
        if (ex[i] == 0) continue;
        if (find(i) != i) continue;
        if (zr[i] == 0) fail(2);
    }
    printf("YES\n");
    return 0;
}
