// 递归实现指数型枚举
vector<int> chosen;
void calc(int x) {
  if (x == n + 1) {
    for (int i = 0; i < chosen.size(); i++)
      printf("%d ", chosen[i]);
    puts("");
    return;
  }
  calc(x + 1);
  chosen.push_back(x);
  calc(x + 1);
  chosen.pop_back();
}

// 递归实现组合型枚举
vector<int> chosen; 
void calc(int x) {
  if (chosen.size() > m || chosen.size() + (n - x + 1) < m) return;
  if (x == n + 1) {
    for (int i = 0; i < chosen.size(); i++)
      printf("%d ", chosen[i]);
    puts("");
    return;
  }
  calc(x + 1);
  chosen.push_back(x);
  calc(x + 1);
  chosen.pop_back();
}

// 递归实现排列型枚举
int order[20];
bool chosen[20];
void calc(int k) {
  if (k == n + 1) {
    for (int i = 1; i <= n; i++)
      printf("%d ", order[i]);
    puts("");
    return;
  }
  for (int i = 1; i <= n; i++) {
    if (chosen[i]) continue;
    order[k] = i;
    chosen[i] = 1;
    calc(k + 1);
    chosen[i] = 0;
    order[k] = 0;
  }
}


// 模拟机器实现，把组合型枚举改为非递归
vector<int> chosen;
int stack[100010], top = 0, address = 0;

void call(int x, int ret_addr) { // 模拟计算机汇编指令call
  int old_top = top;
  stack[++top] = x; // 参数x
  stack[++top] = ret_addr; // 返回地址标号
  stack[++top] = old_top; // 在栈顶记录以前的top值
}

int ret() { // 模拟计算机汇编指令ret
  int ret_addr = stack[top - 1];
  top = stack[top]; // 恢复以前的top值
  return ret_addr;
}

int main() {
  int n, m;
  cin >> n >> m;
  call(1, 0); // calc(1)
  while (top) {
    int x = stack[top - 2]; // 获取参数
    switch (address) {
    case 0:
      if (chosen.size() > m || chosen.size() + (n - x + 1) < m) {
        address = ret(); // return
        continue;
      }
      if (x == n + 1) {
        for (int i = 0; i < chosen.size(); i++)
          printf("%d ", chosen[i]);
        puts("");
        address = ret(); // return
        continue;
      }
      call(x + 1, 1); // 相当于calc(x + 1)，返回后会从case 1继续执行
      address = 0;
      continue; // 回到while循环开头，相当于开始新的递归
    case 1:
      chosen.push_back(x);
      call(x + 1, 2); // 相当于calc(x + 1)，返回后会从case 2继续执行
      address = 0;
      continue; // 回到while循环开头，相当于开始新的递归
    case 2:
      chosen.pop_back();
      address = ret(); // 相当于原calc函数结尾，执行return
    }
  }
}