Problem: In the year 2500 the annual graduation ceremony in the German University in Cairo (GUC) has run smoothly for almost 500 years so far.

The most important part of the ceremony is related to the arrangement of the professors in the ceremonial hall.

Traditionally GUC has n professors. Each professor has his seniority level. All seniorities are different. Let's enumerate the professors from 1 to n, with 1 being the most senior professor and n being the most junior professor.

The ceremonial hall has n seats, one seat for each professor. Some places in this hall are meant for more senior professors than the others. More specifically, m pairs of seats are in "senior-junior" relation, and the tradition requires that for all m pairs of seats (ai, bi) the professor seated in "senior" position ai should be more senior than the professor seated in "junior" position bi.

GUC is very strict about its traditions, which have been carefully observed starting from year 2001. The tradition requires that: 

  * The seating of the professors changes every year. 
  * Year 2001 ceremony was using lexicographically first arrangement of professors in the ceremonial hall. 
  * Each consecutive year lexicographically next arrangement of the professors is used. 



The arrangement of the professors is the list of n integers, where the first integer is the seniority of the professor seated in position number one, the second integer is the seniority of the professor seated in position number two, etc.

Given n, the number of professors, y, the current year and m pairs of restrictions, output the arrangement of the professors for this year.

Input

The first line contains three integers n, y and m (1 ≤ n ≤ 16, 2001 ≤ y ≤ 1018, 0 ≤ m ≤ 100) — the number of professors, the year for which the arrangement should be computed, and the number of pairs of seats for which the seniority relation should be kept, respectively.

The next m lines contain one pair of integers each, "ai bi", indicating that professor on the ai-th seat is more senior than professor on the bi-th seat (1 ≤ ai, bi ≤ n, ai ≠ bi). Some pair may be listed more than once.

Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin stream (you may also use the %I64d specificator).

Output

Print the order in which the professors should be seated in the requested year.

If by this year the GUC would have ran out of arrangements, or the given "senior-junior" relation are contradictory, print "The times have changed" (without quotes).

Examples

Input

3 2001 2
1 2
2 3


Output

1 2 3


Input

7 2020 6
1 2
1 3
2 4
2 5
3 6
3 7


Output

1 2 3 7 4 6 5


Input

10 3630801 0


Output

The times have changed


Input

3 2001 3
1 2
2 3
3 1


Output

The times have changed

Note

In the first example the lexicographically first order of seating is 1 2 3.

In the third example the GUC will run out of arrangements after the year 3630800.

In the fourth example there are no valid arrangements for the seating.

The lexicographical comparison of arrangements is performed by the < operator in modern programming languages. The arrangement a is lexicographically less that the arrangement b, if there exists such i (1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Tpoint {
  double x, y;
  Tpoint() {}
  Tpoint(double _x, double _y) {
    x = _x;
    y = _y;
  }
  inline void read() { scanf("%lf%lf", &x, &y); }
  inline void show() { printf("%lf %lf\n", x, y); }
  inline double norm() { return sqrt(((x) * (x)) + ((y) * (y))); }
};
inline Tpoint operator+(const Tpoint &a, const Tpoint &b) {
  return Tpoint(a.x + b.x, a.y + b.y);
}
inline Tpoint operator-(const Tpoint &a, const Tpoint &b) {
  return Tpoint(a.x - b.x, a.y - b.y);
}
inline Tpoint operator*(const Tpoint &a, const double &b) {
  return Tpoint(a.x * b, a.y * b);
}
inline Tpoint operator/(const Tpoint &a, const double &b) {
  return Tpoint(a.x / b, a.y / b);
}
inline double det(const Tpoint &a, const Tpoint &b) {
  return a.x * b.y - a.y * b.x;
}
inline double dot(const Tpoint &a, const Tpoint &b) {
  return a.x * b.x + a.y * b.y;
}
const long long limit = 1000000000000000000LL;
long long mem[(1 << 16) + 5], y;
int mark[(1 << 16) + 5], stamp;
int n, m, d[20][20];
bool vis[100];
int id[100], must[100], big[100], small[100], ans[100];
inline long long dfs(int step, int mask) {
  if (step < 0) return 1LL;
  long long &res = mem[mask];
  if (mark[mask] == stamp) return res;
  mark[mask] = stamp;
  res = 0;
  int cur = id[step];
  for (int i = 0; i < n; ++i)
    if (!((1 << i) & mask) && (mask & must[i]) == must[i] && cur > big[i] &&
        cur < small[i]) {
      res += dfs(step - 1, mask ^ (1 << i));
      if (res > limit) res = limit + 1;
    }
  return res;
}
inline long long solve() {
  for (int i = 0; i < n; ++i) {
    must[i] = 0;
    small[i] = 100;
    big[i] = -1;
    for (int j = 0; j < n; ++j)
      if (i != j) {
        if (ans[j] != -1) {
          if (d[j][i])
            small[i] = min(small[i], ans[j]);
          else if (d[i][j])
            big[i] = max(big[i], ans[j]);
        } else if (d[j][i])
          must[i] |= 1 << j;
      }
  }
  int st = 0, cnt = 0;
  for (int i = 0; i < n; ++i)
    if (ans[i] != -1) st |= 1 << i;
  for (int i = 0; i < n; ++i)
    if (!vis[i]) id[cnt++] = i;
  ++stamp;
  return dfs(cnt - 1, st);
}
int main() {
  scanf("%d%I64d%d", &n, &y, &m);
  memset(d, 0, sizeof(d));
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    d[a][b] = 1;
  }
  for (int i = 0; i < n; ++i) d[i][i] = 1;
  for (int k = 0; k < n; ++k)
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (d[i][k] && d[k][j]) d[i][j] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      if (d[i][j] && d[j][i]) {
        puts("The times have changed");
        return 0;
      }
  memset(ans, -1, sizeof(ans));
  y -= 2000;
  memset(mark, -1, sizeof(mark));
  memset(vis, false, sizeof(vis));
  stamp = 0;
  if (solve() < y) {
    puts("The times have changed");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = n - 1; j >= 0; --j)
      if (!vis[j]) {
        ans[i] = j;
        vis[j] = true;
        long long t = solve();
        if (t >= y) break;
        y -= t;
        vis[j] = false;
      }
  }
  for (int i = 0; i < n; ++i) {
    if (i) printf(" ");
    printf("%d", n - ans[i]);
  }
  puts("");
  return 0;
}