#AGC059C. [AGC059C] Guessing Permutation for as Long as Possible

[AGC059C] Guessing Permutation for as Long as Possible

Score : 11001100 points

Problem Statement

A teacher has a hidden permutation P=(P1,P2,,PN)P=(P_1,P_2,\ldots,P_N) of (1,2,,N)(1,2,\cdots,N). You are going to determine it.

To do this, you prepared a sequence of pairs of integers $(A_1,B_1),(A_2,B_2),\ldots,(A_{N(N-1)/2},B_{N(N-1)/2})$; this is a permutation of all pairs of the form (a,b)(a,b) (1a<bN1 \le a < b \le N). Now, you will go over the pairs from the beginning. For a pair (Ai,Bi)(A_i, B_i), you will ask if $P_{A_i}, and the teacher will tell you the answer. However, you will skip this question if you can already determine the answer to it from previous answers.

Find the number of permutations PP, for which with your algorithm you will have to ask all N(N1)2\frac{N(N-1)}{2} questions, modulo 109+710^9+7.

Constraints

  • 2N4002 \le N \leq 400
  • 1Ai<BiN1 \le A_i < B_i \le N
  • (Ai,Bi)(Aj,Bj)(A_i,B_i) \neq (A_j,B_j) (iji \neq j)
  • All values in the input are integers.

Input

Input is given from Standard Input in the following format:

NN

A1A_1 B1B_1

A2A_2 B2B_2

\vdots

AN(N1)/2A_{N(N-1)/2} BN(N1)/2B_{N(N-1)/2}

Output

Print the answer.

2
1 2
2

Clearly, for any permutation PP, you need to ask 11 question.

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

Consider P=(2,3,1,4)P=(2,3,1,4) as an example. In this case, you will skip the third question since you know P1<P2P_1 < P_2 and P1>P3P_1 > P_3 from previous questions and you can determine P2>P3P_2 > P_3. Therefore, P=(2,3,1,4)P=(2,3,1,4) should not be counted.

5
1 2
2 3
3 4
4 5
1 5
1 3
2 4
3 5
1 4
2 5
0