CF1842H.Tenzing and Random Real Numbers

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题目描述

There are nn uniform random real variables between 0 and 1, inclusive, which are denoted as x1,x2,,xnx_1, x_2, \ldots, x_n .

Tenzing has mm conditions. Each condition has the form of xi+xj1x_i+x_j\le 1 or xi+xj1x_i+x_j\ge 1 .

Tenzing wants to know the probability that all the conditions are satisfied, modulo 998 244 353998~244~353 .

Formally, let M=998 244 353M = 998~244~353 . It can be shown that the answer can be expressed as an irreducible fraction pq\frac{p}{q} , where pp and qq are integers and q≢0(modM)q \not \equiv 0 \pmod{M} . Output the integer equal to pq1modMp \cdot q^{-1} \bmod M . In other words, output the integer xx that 0x<M0 \le x < M and xqp(modM)x \cdot q \equiv p \pmod{M} .

输入格式

The first line contains two integers nn and mm ( 1n201\le n\le 20 , 0mn2+n0\le m\le n^2+n ).

Then following mm lines of input contains three integers tt , ii and jj ( t{0,1}t \in \{0,1\} , 1ijn1\le i\le j\le n ).

  • If t=0t=0 , the condition is xi+xj1x_i+x_j\le 1 .
  • If t=1t=1 , the condition is xi+xj1x_i+x_j\ge 1 .

It is guaranteed that all conditions are pairwise distinct.

输出格式

Output the probability that all the conditions are satisfied, modulo M=998 244 353M = 998~244~353 .

输入输出样例

  • 输入#1

    3 2
    0 1 2
    1 3 3

    输出#1

    748683265
  • 输入#2

    3 3
    0 1 2
    0 1 3
    0 2 3

    输出#2

    748683265
  • 输入#3

    3 4
    0 1 2
    0 1 3
    1 2 3
    1 2 2

    输出#3

    935854081
  • 输入#4

    4 4
    0 1 2
    0 3 4
    1 1 3
    1 2 4

    输出#4

    0
  • 输入#5

    8 12
    0 1 2
    0 2 3
    1 3 4
    0 1 4
    0 5 6
    0 6 7
    1 7 8
    0 5 8
    1 3 7
    1 3 8
    1 4 7
    1 4 8

    输出#5

    997687297

说明/提示

In the first test case, the conditions are x1+x21x_1+x_2 \le 1 and x3+x31x_3+x_3\ge 1 , and the probability that each condition is satisfied is 12\frac 12 , so the probability that they are both satisfied is 1212=14\frac 12\cdot \frac 12=\frac 14 , modulo 998 244 353998~244~353 is equal to 748683265748683265 .

In the second test case, the answer is 14\frac 14 .

In the third test case, the answer is 116\frac 1{16} .

In the fourth test case, the sum of all variables must equal 22 , so the probability is 00 .

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