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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

typedef long long ll;

ll T[100005];
ll A[100005];

int ans[100005];
ll check[100005];
ll check2[100005];
ll check3[100005];

ll min(ll a, ll b);

ll mod = 1000000007;

int main() {

	int N;

	scanf("%d",&N);

	int i;

	for (i = 0; i < N; i++) {
		scanf(" %lld",&T[i]);
	}
	for (i = 0; i < N; i++) {
		scanf(" %lld", &A[i]);
	}

	memset(ans,-1,sizeof(ans));

	ans[0] = (int)T[0];
	ans[N - 1] = (int)A[N - 1];

	for (i = 0; i < N-1; i++) {
		if (T[i] != T[i + 1]) {
			ans[i + 1] = (int)T[i + 1];
		}
	}
	for (i = N - 1; i >= 1; i--) {
		if (A[i] != A[i - 1]) {
			ans[i - 1] = (int)A[i - 1];
		}
	}
	ll answer = 1ll;
	for (i = 0; i < N; i++) {
		if (ans[i] == -1) { 
			answer *= min(T[i], A[i]); 
			answer %= mod;
			check[i] = min(T[i],A[i]);
		}
		else {
			check[i] = (ll)ans[i];
		}
	}
	ll ma = -1ll;

	for (i = 0; i < N; i++) {
		if (ma < check[i]) {
			ma = check[i];
		}
		check2[i] = ma;
	}
	ma = -1ll;
	for (i = N-1; i >= 0; i--) {
		if (ma < check[i]) {
			ma = check[i];
		}
		check3[i] = ma;
	}

	int flag = 0;

	for (i = 0; i < N; i++) {
		if (check2[i] != T[i] || check3[i] != A[i]) { 
			flag = 1;
			break;
		}
	}
	if (flag == 0) { printf("%lld\n",answer); }
	else { printf("0\n"); }

	return 0;
}
ll min(ll a, ll b) {
	if (a >= b) { return b; }
	else { return a; }
}