Problem: Help! A robot escaped our lab and we need help finding it. 

The lab is at the point (0, 0) of the coordinate plane, at time 0 the robot was there. The robot's movements are defined by a program — a string of length l, consisting of characters U, L, D, R. Each second the robot executes the next command in his program: if the current coordinates of the robot are (x, y), then commands U, L, D, R move it to cells (x, y + 1), (x - 1, y), (x, y - 1), (x + 1, y) respectively. The execution of the program started at time 0. The program is looped, i.e. each l seconds of executing the program start again from the first character. Unfortunately, we don't know what program was loaded into the robot when he left the lab.

Our radars managed to find out the position of the robot at n moments of time: we know that at the moment of time ti the robot is at the point (xi, yi). Given this data, either help to determine what program could be loaded into the robot, or determine that no possible program meets the data and the robot must have broken down.

Input

The first line of the input contains two space-separated integers n and l (1 ≤ n ≤ 2·105, 1 ≤ l ≤ 2·106).

Next n lines contain three space-separated integers — ti, xi, yi (1 ≤ ti ≤ 1018,  - 1018 ≤ xi, yi ≤ 1018). The radar data is given chronologically, i.e. ti < ti + 1 for all i from 1 to n - 1.

Output

Print any of the possible programs that meet the data. If no program meets the data, print a single word 'NO' (without the quotes).

Examples

Input

3 3
1 1 0
2 1 -1
3 0 -1


Output

RDL


Input

2 2
1 1 0
999 1 0


Output

RL


Input

2 5
10 10 0
20 0 0


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
const int maxn = 200000;
const int __buffsize = 100000;
char __buff[__buffsize];
char *__buffs, *__buffe;
template <typename T>
inline T &Read(T &first) {
  static char c;
  while (1) {
    c = (__buffs == __buffe ? fread(__buff, 1, __buffsize, stdin),
         __buffe = __buff + __buffsize, *((__buffs = __buff)++) : *(__buffs++));
    if (c == '-' || (c >= '0' && c <= '9')) break;
  }
  bool flag = c == '-';
  first = flag ? 0 : c - '0';
  while (1) {
    c = (__buffs == __buffe ? fread(__buff, 1, __buffsize, stdin),
         __buffe = __buff + __buffsize, *((__buffs = __buff)++) : *(__buffs++));
    if (c < '0' || c > '9') break;
    (first *= 10) += c - '0';
  }
  if (flag) first = -first;
  return first;
}
struct data {
  int t;
  long long first, second, z;
  data() {}
  data(int _t, long long _x, long long _y, const long long &_z)
      : t(_t), first(_x), second(_y), z(_z) {}
  friend bool operator<(const data &first, const data &second) {
    return first.t < second.t;
  }
};
data a[maxn + 5];
int n, l;
long long T[maxn + 5], X[maxn + 5], Y[maxn + 5];
inline long long Floor(long long a, long long b);
inline long long Ceil(long long a, long long b);
inline long long Floor(long long a, long long b) {
  if (b < 0) a = -a, b = -b;
  if (a < 0) return -Ceil(-a, b);
  return a / b;
}
inline long long Ceil(long long a, long long b) {
  if (b < 0) a = -a, b = -b;
  if (a < 0) return -Floor(-a, b);
  return (a + b - 1) / b;
}
int main() {
  Read(n), Read(l);
  for (int i = (0), _end_ = (n); i < _end_; ++i) {
    Read(T[i]), Read(X[i]), Read(Y[i]);
    if ((T[i] ^ X[i] ^ Y[i]) & 1) printf("NO\n"), exit(0);
    a[i] = data(T[i] % l, X[i], Y[i], T[i] / l);
  }
  a[n++] = data(0, 0, 0, 0);
  a[n++] = data(l, 0, 0, -1);
  sort(a, a + n);
  long long Lx = LLONG_MIN, Rx = LLONG_MAX, Ly = LLONG_MIN, Ry = LLONG_MAX;
  for (int i = (1), _end_ = (n); i < _end_; ++i) {
    if (a[i].z == a[i - 1].z) {
      if (abs(a[i].first - a[i - 1].first) +
              abs(a[i].second - a[i - 1].second) >
          (a[i].t - a[i - 1].t))
        printf("NO\n"), exit(0);
    } else {
      long long dx = a[i].first - a[i - 1].first,
                dy = a[i].second - a[i - 1].second;
      long long dz = a[i].z - a[i - 1].z;
      long long tmpx = dx - dy, tmpy = dx + dy;
      int dt = a[i].t - a[i - 1].t;
      long long lx = tmpx - dt, rx = tmpx + dt, ly = tmpy - dt, ry = tmpy + dt;
      if (dz < 0) swap(lx, rx), swap(ly, ry);
      lx = Ceil(lx, dz), rx = Floor(rx, dz), ly = Ceil(ly, dz),
      ry = Floor(ry, dz);
      chkmax(Lx, lx);
      chkmin(Rx, rx);
      chkmax(Ly, ly);
      chkmin(Ry, ry);
    }
  }
  if (Lx > Rx || Ly > Ry) printf("NO"), exit(0);
  bool flag = 0;
  long long ansx = 0, ansy = 0;
  for (long long i = Lx; i < Lx + 4; ++i)
    for (long long j = Ly; j < Ly + 4; ++j)
      if (i <= Rx && j <= Ry && !((i + j) & 1)) {
        long long ux = (j + i) >> 1, uy = (j - i) >> 1;
        if (!((ux + uy + l) & 1)) {
          flag = 1;
          ansx = ux, ansy = uy;
        }
      }
  if (!flag) printf("NO"), exit(0);
  for (int i = (0), _end_ = (n); i < _end_; ++i)
    a[i].first -= ansx * a[i].z, a[i].second -= ansy * a[i].z;
  a[n++] = data(l, ansx, ansy, 0);
  int curx = 0, cury = 0, curt = 0;
  for (int i = (0), _end_ = (n); i < _end_; ++i) {
    while (curx < a[i].first) putchar('R'), ++curx, ++curt;
    while (curx > a[i].first) putchar('L'), --curx, ++curt;
    while (cury < a[i].second) putchar('U'), ++cury, ++curt;
    while (cury > a[i].second) putchar('D'), --cury, ++curt;
    while (curt < a[i].t) putchar('L'), putchar('R'), curt += 2;
  }
  return 0;
}