Problem: Yet another education system reform has been carried out in Berland recently. The innovations are as follows:

An academic year now consists of n days. Each day pupils study exactly one of m subjects, besides, each subject is studied for no more than one day. After the lessons of the i-th subject pupils get the home task that contains no less than ai and no more than bi exercises. Besides, each subject has a special attribute, the complexity (ci). A school can make its own timetable, considering the following conditions are satisfied:

  * the timetable should contain the subjects in the order of the complexity's strict increasing; 
  * each day, except for the first one, the task should contain either k times more exercises, or more by k compared to the previous day (more formally: let's call the number of home task exercises in the i-th day as xi, then for each i (1 < i ≤ n): either xi = k + xi - 1 or xi = k·xi - 1 must be true); 
  * the total number of exercises in all home tasks should be maximal possible. 



All limitations are separately set for each school.

It turned out that in many cases ai and bi reach 1016 (however, as the Berland Minister of Education is famous for his love to half-measures, the value of bi - ai doesn't exceed 100). That also happened in the Berland School №256. Nevertheless, you as the school's principal still have to work out the timetable for the next academic year...

Input

The first line contains three integers n, m, k (1 ≤ n ≤ m ≤ 50, 1 ≤ k ≤ 100) which represent the number of days in an academic year, the number of subjects and the k parameter correspondingly. Each of the following m lines contains the description of a subject as three integers ai, bi, ci (1 ≤ ai ≤ bi ≤ 1016, bi - ai ≤ 100, 1 ≤ ci ≤ 100) — two limitations to the number of exercises on the i-th subject and the complexity of the i-th subject, correspondingly. Distinct subjects can have the same complexity. The subjects are numbered with integers from 1 to m. 

Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin stream or the %I64d specificator.

Output

If no valid solution exists, print the single word "NO" (without the quotes). Otherwise, the first line should contain the word "YES" (without the quotes) and the next n lines should contain any timetable that satisfies all the conditions. The i + 1-th line should contain two positive integers: the number of the subject to study on the i-th day and the number of home task exercises given for this subject. The timetable should contain exactly n subjects.

Examples

Input

4 5 2
1 10 1
1 10 2
1 10 3
1 20 4
1 100 5


Output

YES
2 8
3 10
4 20
5 40


Input

3 4 3
1 3 1
2 4 4
2 3 3
2 2 2


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
long long dp[52][52][102];
struct node1 {
  int i, k;
} pre[52][52][102];
struct node {
  long long a, b;
  int c;
  int idx;
  bool operator<(const node &aa) const { return c < aa.c; }
} ex[55];
void print(int i, int j, int k) {
  if (j > 1) print(pre[i][j][k].i, j - 1, pre[i][j][k].k);
  printf("%d %I64d\n", ex[i].idx, ex[i].a + k);
}
long long _sum;
void _func(int i, int j, int k) {
  _sum += dp[i][j][k];
  if (j > 1) _func(pre[i][j][k].i, j - 1, pre[i][j][k].k);
}
int main() {
  int n, m, K;
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= m; i++) {
    scanf("%I64d%I64d%d", &ex[i].a, &ex[i].b, &ex[i].c);
    ex[i].idx = i;
  }
  sort(ex + 1, ex + m + 1);
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) {
      long long end = ex[i].b - ex[i].a;
      for (int k = 0; k <= end; k++) {
        if (j == 1)
          dp[i][j][k] = ex[i].a + k;
        else {
          long long _max = 0, t = k + ex[i].a;
          int j1 = j - 1;
          for (int i1 = i - 1; i1 > 0; i1--) {
            if (ex[i1].c >= ex[i].c) continue;
            long long k1 = t - K;
            if (k1 >= ex[i1].a && k1 <= ex[i1].b) {
              if (_max < dp[i1][j1][k1 - ex[i1].a]) {
                _max = dp[i1][j1][k1 - ex[i1].a];
                pre[i][j][k].i = i1;
                pre[i][j][k].k = k1 - ex[i1].a;
              }
            }
            k1 = t / K;
            if (k1 * K == t && k1 >= ex[i1].a && k1 <= ex[i1].b) {
              if (_max < dp[i1][j1][k1 - ex[i1].a]) {
                _max = dp[i1][j1][k1 - ex[i1].a];
                pre[i][j][k].i = i1;
                pre[i][j][k].k = k1 - ex[i1].a;
              }
            }
          }
          if (_max != 0) dp[i][j][k] = _max + t;
        }
      }
    }
  long long _max = 0;
  int i1, k1;
  for (int i = n; i <= m; i++) {
    long long end = ex[i].b - ex[i].a;
    for (int k = 0; k <= end; k++) {
      if (_max < dp[i][n][k]) {
        _max = dp[i][n][k];
        i1 = i;
        k1 = k;
      }
    }
  }
  if (_max == 0)
    printf("NO\n");
  else {
    printf("YES\n");
    print(i1, n, k1);
  }
  return 0;
}