Anton likes to play chess. Also, he likes to do programming. That is why he decided to write the program that plays chess. However, he finds the game on 8 to 8 board to too simple, he uses an infinite one instead.

The first task he faced is to check whether the king is in check. Anton doesn't know how to implement this so he asks you to help.

Consider that an infinite chess board contains one white king and the number of black pieces. There are only rooks, bishops and queens, as the other pieces are not supported yet. The white king is said to be in check if at least one black piece can reach the cell with the king in one move. 

Help Anton and write the program that for the given position determines whether the white king is in check.

Remainder, on how do chess pieces move: 

  * Bishop moves any number of cells diagonally, but it can't "leap" over the occupied cells. 
  * Rook moves any number of cells horizontally or vertically, but it also can't "leap" over the occupied cells. 
  * Queen is able to move any number of cells horizontally, vertically or diagonally, but it also can't "leap". 

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 500 000) — the number of black pieces.

The second line contains two integers x0 and y0 ( - 109 ≤ x0, y0 ≤ 109) — coordinates of the white king.

Then follow n lines, each of them contains a character and two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — type of the i-th piece and its position. Character 'B' stands for the bishop, 'R' for the rook and 'Q' for the queen. It's guaranteed that no two pieces occupy the same position.

Output

The only line of the output should contains "YES" (without quotes) if the white king is in check and "NO" (without quotes) otherwise.

Examples

Input

2
4 2
R 1 1
B 1 5


Output

YES


Input

2
4 2
R 3 3
B 1 5


Output

NO

Note

Picture for the first sample: 

<image> White king is in check, because the black bishop can reach the cell with the white king in one move. The answer is "YES".

Picture for the second sample: 

<image> Here bishop can't reach the cell with the white king, because his path is blocked by the rook, and the bishop cant "leap" over it. Rook can't reach the white king, because it can't move diagonally. Hence, the king is not in check and the answer is "NO".
import java.awt.*;
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.NoSuchElementException;
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Supplier;

public class Main {
  void run() {
    int n = ni();
    int x0 = ni();
    int y0 = ni();

    Point atom = new Point(x0, y0);
    Point[] closest = new Point[8];
    char[] closest_type = new char[8];
    int[][] ofs = {
        {0, 1, 2},
        {3, -1, 4},
        {5, 6, 7}
    };
    for (int i = 0; i < n; ++i) {
      char t = sc.next().charAt(0);
      int x = ni();
      int y = ni();
      if (Math.abs(x - x0) == Math.abs(y - y0) || x == x0 || y == y0) {
        int dx = x == x0 ? 1 : x > x0 ? 2 : 0;
        int dy = y == y0 ? 1 : y > y0 ? 2 : 0;
        int idx = ofs[dy][dx];
        Point trg = new Point(x, y);
        if (closest[idx] == null || closest[idx].distance(atom) > trg.distance(atom)) {
          closest[idx] = trg;
          closest_type[idx] = t;
        }
      }
    }
    boolean flag = false;
    for (int i = 0; i < 8; ++i) {
      flag |= closest_type[i] == 'Q';
      if (i == 0 || i == 2 || i == 5 || i == 7) {
        flag |= closest_type[i] == 'B';
      } else {
        flag |= closest_type[i] == 'R';
      }
    }

    System.out.println(flag ? "YES" : "NO");
  }

  FastScanner sc = new FastScanner(System.in);

  public static void main(String[] args) {
    new Main().run();
  }

  int ni() {
    return Integer.parseInt(sc.next());
  }

  void debug(Object... os) {
    System.err.println(Arrays.deepToString(os));
  }

  class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    /**
     * 1-indexed なBinary Indexed Treeを構築する
     *
     * @param n   容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
      this.n = n;
      bit = new ArrayList<>(n + 1);
      for (int i = 0; i < n + 1; ++i) {
        bit.add(sup.get());
      }
      this.bif = bif;
    }

    /**
     * iの位置の値をvで更新する
     *
     * @param i index
     * @param v 新しい値
     */
    void set(int i, T v) {
      for (int x = i; x <= n; x += x & -x) {
        bit.set(x, bif.apply(bit.get(x), v));
      }
    }

    /**
     * クエリー
     *
     * @param defaultValue 初期値
     * @param i            index
     * @return [1, i]までfを適用した結果
     */
    T reduce(T defaultValue, int i) {
      T ret = defaultValue;
      for (int x = i; x > 0; x -= x & -x) {
        ret = bif.apply(ret, bit.get(x));
      }
      return ret;
    }
  }

  class SegmentTree<T> {
    int n;
    ArrayList<T> dat;
    BiFunction<T, T, T> bif;
    Supplier<T> sup;

    /**
     * 0-indexed なSegment Treeを構築する
     *
     * @param n_  要求容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    SegmentTree(int n_, BiFunction<T, T, T> bif, Supplier<T> sup) {
      n = 1;
      while (n < n_) n *= 2;

      dat = new ArrayList<>(2 * n - 1);
      for (int i = 0; i < 2 * n - 1; ++i) {
        dat.add(sup.get());
      }
      this.bif = bif;
      this.sup = sup;
    }

    /**
     * kの位置の値をvで更新する
     *
     * @param k index
     * @param v 新しい値
     */
    void set(int k, T v) {
      k += n - 1;
      dat.set(k, v);
      while (k > 0) {
        k = (k - 1) / 2;
        dat.set(k, bif.apply(dat.get(k * 2 + 1), dat.get(k * 2 + 2)));
      }
    }

    /**
     * クエリー
     *
     * @param l はじめ
     * @param r おわり
     * @return [l, r)での演算bifを適用した結果を返す
     */
    T reduce(int l, int r) {
      return _reduce(l, r, 0, 0, n);
    }

    T _reduce(int a, int b, int k, int l, int r) {
      if (r <= a || b <= l) return sup.get();
      if (a <= l && r <= b) return dat.get(k);
      T vl = _reduce(a, b, k * 2 + 1, l, (l + r) / 2);
      T vr = _reduce(a, b, k * 2 + 2, (l + r) / 2, r);
      return bif.apply(vl, vr);
    }
  }

  class Pair<F extends Comparable<F>, S extends Comparable<S>> implements Comparable<Pair<F, S>> {
    F f;
    S s;

    Pair() {
    }

    Pair(F f, S s) {
      this.f = f;
      this.s = s;
    }

    Pair(Pair<F, S> p) {
      f = p.f;
      s = p.s;
    }

    @Override
    public int compareTo(Pair<F, S> p) {
      if (f.compareTo(p.f) != 0) {
        return f.compareTo(p.f);
      }
      return s.compareTo(p.s);
    }

    @Override
    public int hashCode() {
      return f.hashCode() ^ s.hashCode();
    }

    @Override
    public boolean equals(Object o) {
      if (this == o) {
        return true;
      }
      if (o == null || this.f == null || this.s == null) {
        return false;
      }
      if (this.getClass() != o.getClass()) {
        return false;
      }
      Pair p = (Pair) o;
      return this.f.equals(p.f) && this.s.equals(p.s);
    }

    @Override
    public String toString() {
      return "{" + f.toString() + ", " + s.toString() + "}";
    }
  }

  /**
   * ユークリッドの互除法
   *
   * @return a と b の最大公約数
   */
  long gcd(long a, long b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }

  /**
   * 拡張ユークリッドの互除法
   *
   * @return mx + ny = gcd(m, n)となるような(x, y)を返す
   */
  Pair<Long, Long> gcd_ex(long m, long n) {
    long[][] mat = _gcd_ex(m, n);
    return new Pair<>(mat[0][0], mat[0][1]);
  }

  long[][] _gcd_ex(long m, long n) {
    if (n == 0) {
      return new long[][]{{1, 0}, {0, 1}};
    }
    long k = m / n;
    long[][] K = new long[][]{{0, 1}, {1, -k}};
    long[][] r = _gcd_ex(n, m % n);
    long[][] dst = new long[2][2];
    for (int y = 0; y < 2; ++y)
      for (int x = 0; x < 2; ++x)
        for (int i = 0; i < 2; ++i)
          dst[y][x] += r[y][i] * K[i][x];
    return dst;
  }

  long MOD = 1_000_000_007;

  /**
   * 繰り返し2乗法を用いたべき乗の実装
   *
   * @return a^r (mod 1,000,000,007)
   */
  long pow(long a, long r) {
    long sum = 1;
    while (r > 0) {
      if ((r & 1) == 1) {
        sum *= a;
        sum %= MOD;
      }
      a *= a;
      a %= MOD;
      r >>= 1;
    }
    return sum;
  }

  /**
   * 組み合わせ
   * O(n)
   *
   * @return {}_nC_r
   */
  long C(int n, int r) {
    long sum = 1;
    for (int i = n; 0 < i; --i) {
      sum *= i;
      sum %= MOD;
    }
    long s = 1;
    for (int i = r; 0 < i; --i) {
      s *= i;
      s %= MOD;
    }
    sum *= pow(s, MOD - 2);
    sum %= MOD;

    long t = 1;
    for (int i = n - r; 0 < i; --i) {
      t *= i;
      t %= MOD;
    }
    sum *= pow(t, MOD - 2);
    sum %= MOD;

    return sum;
  }

  double GOLDEN_RATIO = (1.0 + Math.sqrt(5)) / 2.0;

  /**
   * 黄金分割探索
   *
   * @param left  下限
   * @param right 上限
   * @param f     探索する関数
   * @param comp  上に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue)
   *              下に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue).reversed()
   * @return 極値の座標x
   */
  double goldenSectionSearch(double left, double right, Function<Double, Double> f, Comparator<Double> comp) {
    double c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
    double c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
    double d1 = f.apply(c1);
    double d2 = f.apply(c2);
    while (right - left > 1e-9) {
      if (comp.compare(d1, d2) > 0) {
        right = c2;
        c2 = c1;
        d2 = d1;
        c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
        d1 = f.apply(c1);
      } else {
        left = c1;
        c1 = c2;
        d1 = d2;
        c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
        d2 = f.apply(c2);
      }
    }
    return right;
  }

  /**
   * [a,b]をm:nに内分する点を返す
   */
  double divideInternally(double a, double b, double m, double n) {
    return (n * a + m * b) / (m + n);
  }

  /**
   * http://qiita.com/p_shiki37/items/65c18f88f4d24b2c528b
   */
  static class FastScanner {
    private final InputStream in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    public FastScanner(InputStream in) {
      this.in = in;
    }

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) return buffer[ptr++];
      else return -1;
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
      skipUnprintable();
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }
  }
}