#ABC273H. [ABC273Ex] Inv(0,1)ving Insert(1,0)n

[ABC273Ex] Inv(0,1)ving Insert(1,0)n

Score : 600600 points

Problem Statement

We have a sequence AA consisting of integer pairs. Initially, A=((0,1),(1,0))A = ( (0, 1), (1, 0) ).

You may perform the following operation on AA as many (possibly zero) times as you want:

  • choose adjacent two integer pairs (a,b)(a, b) and (c,d)(c, d), and insert (a+c,b+d)(a + c, b + d) between them.

For a sequence TT consisting of integer pairs, let us define f(T)f(T) as follows:

  • let f(T)=f(T) = (the minimum number of operations required to make every element of TT contained in AA).- We say that "every element of TT is contained in AA" if, for all elements xx contained in TT, xx is contained in (the set consisting of the elements contained in AA).
  • We say that "every element of TT is contained in AA" if, for all elements xx contained in TT, xx is contained in (the set consisting of the elements contained in AA).
  • Here, if there are no such operations, let f(T)=0f(T) = 0.

There is a sequence S=((a1,b1),(a2,b2),,(aN,bN))S = ((a_1, b_1), (a_2, b_2), \dots, (a_N, b_N)) consisting of NN integer pairs. Here, all elements of SS are pairwise distinct. There are N×(N+1)2\frac{N \times (N+1)}{2} possible consecutive subarrays $S_{l,r}=((a_l,b_l),(a_{l+1},b_{l+1}),\dots,(a_r,b_r))$ of SS. Find the sum, modulo 998244353998244353, of f(Sl,r)f(S_{l,r}) over those subarrays. Formally, find $\displaystyle \sum^{N} _ {l=1} \sum^{N} _ {r=l} f(S_{l,r})$, modulo 998244353998244353.

Constraints

  • 1N1051 \le N \le 10^5
  • 0ai,bi1090 \le a_i,b_i \le 10^9
  • aiaja_i \neq a_j or bibjb_i \neq b_j, if iji \neq j.

Input

The input is given from Standard Input in the following format:

NN

a1a_1 b1b_1

a2a_2 b2b_2

\dots

aNa_N bNb_N

Output

Print the answer as an integer.

7
1 2
3 7
3 5
0 0
1000000000 1
0 1
6 3
3511324
  • f(S1,1)=2f(S_{1,1})=2.- We can make $((0,1),(1,0)) \rightarrow ((0,1),(1,1),(1,0)) \rightarrow ((0,1),(1,2),(1,1),(1,0))$.
  • We can make $((0,1),(1,0)) \rightarrow ((0,1),(1,1),(1,0)) \rightarrow ((0,1),(1,2),(1,1),(1,0))$.
  • f(S1,2)=5f(S_{1,2})=5.
  • f(S1,3)=7f(S_{1,3})=7.
  • f(S2,2)=5f(S_{2,2})=5.
  • f(S2,3)=7f(S_{2,3})=7.
  • f(S3,3)=4f(S_{3,3})=4.
  • f(S5,5)=1000000000=109f(S_{5,5})=1000000000 = 10^9.
  • f(S5,6)=1000000000=109f(S_{5,6})=1000000000 = 10^9.
  • f(S6,6)=0f(S_{6,6})=0.- (0,1)(0, 1) is originally contained in AA.
  • (0,1)(0, 1) is originally contained in AA.
  • f(Sl,r)=0f(S_{l,r})=0 for all Sl,rS_{l,r} not mentioned above.- We can prove that AA can never contain (0,0)(0,0) or (6,3)(6,3) regardless of operations.
  • We can prove that AA can never contain (0,0)(0,0) or (6,3)(6,3) regardless of operations.

Therefore, the sum of f(Sl,r)f(S_{l,r}) is 2000000030=2×109+302000000030 = 2 \times 10^9 + 30, whose remainder when divided by 998244353998244353 is 35113243511324.