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;

#define int long long

#define inf 1000000007LL
#define mod 1000000007LL

#define rep(i, n) for(int i = 0; i < (n); i++)
#define trep(i, n) for(int i = 0; i <= (n); i++)
#define rrep(i, n) for(int i = (n) - 1; i >= 0; i--)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define mfor(i, s, t) for(int i = (s); i < (t); i++)
#define tfor(i, s, t) for(int i = (s); i <= (t); i++)
#define rfor(i, s, t) for(int i = (t) - 1; i >= (s); i--)

int k[514];
bool ok[514];
vector<int> v[514];

bool okok[114514];
pair<int, int> f[114514];

void dfs(int p) {
  if(!ok[p]) {
    ok[p] = true;
    for(auto i : v[p]) {
      okok[i] = true;
      dfs(f[i].first);
      dfs(f[i].second);
    }
  }
}

signed main() {
  int n, h;
  cin >> n >> h;
  rep(i, (h + 10) * 2) {
    k[i] = 0;
    ok[i] = false;
  }
  rep(i, n) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if(c == 0) {
      v[a].push_back(i);
      k[a]++;
      f[i].first = a;
    }
    else {
      v[c + h + 10].push_back(i);
      k[c + h + 10]--;
      f[i].first = c + h + 10;
    }
    if(d == 0) {
      v[b + h + 10].push_back(i);
      k[b + h + 10]++;
      f[i].second = b + h + 10;
    }
    else {
      v[d].push_back(i);
      k[d]--;
      f[i].second = d;
    }
  }
  rep(i, (h + 10) * 2) {
    if(k[i] < 0) {
      cout << "NO" << endl;
      return 0;
    }
  }
  rep(i, n) {
    okok[i] = false;
  }
  rep(i, (h + 10) * 2) {
    if(k[i] > 0) {
      dfs(i);
    }
  }
  rep(i, n) {
    if(!okok[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
}
