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.

Examples

Input

5
1 3 3 3 3
3 3 2 2 2


Output

4


Input

5
1 1 1 2 2
3 2 1 1 1


Output

0


Input

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


Output

884111967


Input

1
17
17


Output

1
#include<bits/stdc++.h>
using namespace std;

#define rep(i, n) for(int (i)=0; (i)<(n); (i)++) 
using LL = long long;

const LL MOD = 1e9+7;
 
int N;
int A[114514];
int T[114514];
int L[114514];
int R[114514];

int main() {
  scanf("%d", &N);
  rep(i, N) {
    scanf("%d", &T[i+1]);
  }
  rep(i, N) {
    scanf("%d", &A[i+1]);
  }
  rep(i, N) {
    if (T[i] < T[i+1]) {
      R[i+1] = L[i+1] = T[i+1];
    } else {
      L[i+1] = 1;
      R[i+1] = T[i+1];
    }
  }
  LL ans = 1;
  rep(i, N) {
    int j=i+1;
    if (A[j] > A[j+1]) {
      L[j] = max(L[j], A[j]);
      R[j] = min(R[j], A[j]);
    } else {
      L[j] = max(L[j], 1);
      R[j] = min(R[j], A[j]);
    }
    
    if (L[j] > R[j]) {
      puts("0");
      return 0;
    }
    (ans *= LL(R[j]-L[j]+1)) %= MOD;
  }
  printf("%lld\n", ans);
}
      
    
    
