Problem:
Score : 400 points

Problem Statement

Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range.
The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N.
Mr. Takahashi traversed the range from the west and Mr. Aoki from the east.

The height of Mt. i is h_i, but they have forgotten the value of each h_i.
Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i).
Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i.

We know that the height of each mountain h_i is a positive integer.
Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7.

Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights.
In such a case, output 0.

Constraints

1 ≤ N ≤ 10^5

1 ≤ T_i ≤ 10^9

1 ≤ A_i ≤ 10^9

T_i ≤ T_{i+1} (1 ≤ i ≤ N - 1)

A_i ≥ A_{i+1} (1 ≤ i ≤ N - 1)

Input

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

N
T_1 T_2 ... T_N
A_1 A_2 ... A_N

Output

Print the number of possible sequences of the mountains' heights, modulo 10^9 + 7.

Sample Input 1

5
1 3 3 3 3
3 3 2 2 2

Sample Output 1

4

The possible sequences of the mountains' heights are:

1, 3, 2, 2, 2

1, 3, 2, 1, 2

1, 3, 1, 2, 2

1, 3, 1, 1, 2

for a total of four sequences.

Sample Input 2

5
1 1 1 2 2
3 2 1 1 1

Sample Output 2

0

The records are contradictory, since Mr. Takahashi recorded 2 as the highest peak after climbing all the mountains but Mr. Aoki recorded 3.

Sample Input 3

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5

Sample Output 3

884111967

Don't forget to compute the number modulo 10^9 + 7.

Sample Input 4

1
17
17

Sample Output 4

1

Some mountain ranges consist of only one mountain.

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

--記録の無矛盾条件は、
--Tが単調増加、Aが単調減少で,
--かつ、両者が一致するところでは両者の最大値をとることである。
--単調性は保証されているので、エラーの検出は、両者の値が一致した時に、それが最大なことを見る。
--最高高さ記録のタプルの要素に対して、これが正しいとき、
--記録の大きい側は参考にならないので無視する。
--記録の小さい方について、0から初めて、真に値が増加するとき、その高さはその値であるので、場合の数は×１である。
--小さい方の記録が変化しないとき、その山は今までの記録だけの可能性があるので、場合の数は×記録値である。

divider::[(Int,Int)]->([(Int,Int)],[(Int,Int)],[(Int,Int)])
divider [] = ([],[],[])
divider xs =  let (a,ys) = (span (\(t,a)->t <a) xs) in
              let (b,c ) = (span (\(t,a)->t==a) ys) in (a,b,c)


--最初のt<a区間：エラー検出は、aが変化する時。記録は１以上なので、初期値は０だとして、これで先頭を処理する。
solver1::[(Int,Int)]->(Int,Int)-> Int
solver1 []         _             = 1
solver1 ((t,a):xs) (old_t,old_a) = if (old_a /=0)&&(old_a /= a) then 0
                                                                else if t > old_t then 1 * (solver1 xs (t,a))
                                                                                  else mod (t * (solver1 xs (t,a))) 1000000007
--一致区間
solver2::[(Int,Int)]->Int
solovr2 []        =0
solver2 (_:[])    =1
solver2 (_:_:[])  =1
solver2 ((t,a):xs)= if t==a then mod (t * (solver2 xs) ) 1000000007 else 0

solver3::[(Int,Int)]->(Int,Int)->Int
solver3 dat  (t,a) =solver1 (reverse (map (\(t,a)->(a,t)) dat)) (t,a)

solver_tot::[(Int,Int)]->Int
solver_tot xs = let (dat1,dat2,dat3) = divider xs in mod ((solver1 dat1 (0,0))*(solver2 dat2)*(solver3 dat3 (0,0))) 1000000007

main::IO()
main = do
  _<-getLine
  ts<-getLine
  as<-getLine
  let dat = (zip  (map read (words ts)) (map read (words as))) :: [(Int,Int)]
  print (solver_tot dat)