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 <algorithm>
#include <cassert>
#include <cstring>
#include <iostream>
#include <vector>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define TRACE(x) cout << #x << " = " << x << endl
#define _ << " _ " <<

typedef long long llint;

const int MAX = 444;

vector<int> E[MAX];
bool bio[MAX];

int inDeg[MAX];
int outDeg[MAX];
int N, H;

vector<int> v;

void dfs(int x) {
  if (bio[x]) return;
  bio[x] = true;
  v.push_back(x);
  for (int y: E[x]) dfs(y);
}

bool solve() {
  REP(i, 2*H) {
    if (!bio[i] && E[i].size()) {
      v.clear();
      dfs(i);
      
      bool haveStart = false;
      bool haveEnd = false;
      
      int balance = 0;
      for (int x: v) {
        if (x < H) {
          haveStart |= inDeg[x] < outDeg[x];
          if (inDeg[x] > outDeg[x]) return false;
          balance += outDeg[x] - inDeg[x];
        } else {
          haveEnd |= inDeg[x] > outDeg[x];
          if (inDeg[x] < outDeg[x]) return false;
          balance -= inDeg[x] - outDeg[x];
        }
      }
      if (!haveStart || !haveEnd || balance != 0) return false;
    }
  }
  return true;
}

int main(void) {
  scanf("%d %d", &N, &H);
  REP(i, N) {
    int A, B, C, D;
    scanf("%d %d %d %d", &A, &B, &C, &D);

    int u = C == 0 ? A-1 : H+C-1;
    int v = D == 0 ? H+B-1 : D-1;

    E[u].push_back(v);
    E[v].push_back(u);
    outDeg[u]++;
    inDeg[v]++;
  }

  puts(solve() ? "YES" : "NO");
  return 0;
}
