#include <bits/stdc++.h>
using namespace std;
#define dbg(x) x
#define prt(x) dbg(cerr << x)
#define pv(x) dbg(cerr << #x << " = " << x << '\n')
#define parr(x) dbg(cerr << #x << " = "; for (auto y : x) cerr << y << ' '; cerr << '\n';)
#define parr2d(x) dbg(cerr << #x << " = \n"; for (auto _ : x) {parr(_);} cerr << '\n';)
/*
1
bitmask
2
for each pair of walls
calculate the sum of all possible sums of elements between these
such that all of them are less than a certain value
for this, doing all ranges & all values will work
so now, what if there's an even taller segment on the outside?
in addition to calculating the segment's sum, also calculate the # of
ways for things on the outside to exist such that it's not true that
there's a bigger element to the left AND the right
ways_left * smaller_right + smaller_left * ways_right
- smaller_left * smaller_right
3
4
5
let i be the end of a segment
dp[i] = the answer
ways[i] = # of ways so that h[i] = 2
for i, for all j <= i - 2
dp[i] += dp[j] + ways[j] * (i - j - 1)
ways[i] += ways[j]
*/
const long long mod = 1e9 + 7;
int main () {
ios::sync_with_stdio(0); cin.tie(0);
int n;
cin >> n;
vector<int> a(n), b(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < n; i++) {
cin >> b[i];
}
long long ans = 0;
for (int mask = 0; mask < (1 << n); mask++) {
vector<int> c(n);
for (int i = 0; i < n; i++) {
c[i] = ((mask & (1 << i)) ? b[i] : a[i]);
}
int mxe = 0, ind = 0;
for (int i = 0; i < n; i++) {
if (c[i] > mxe) {
mxe = c[i];
ind = i;
}
}
long long sum = 0, mx = 0;
int prev = -1;
for (int i = 0; i <= ind; i++) {
if (c[i] >= mx) {
if (i > 0) {
ans += ((long long) (i - prev - 1) * mx - sum) % mod; ans %= mod;
}
mx = c[i];
prev = i;
sum = 0;
} else {
sum += c[i];
}
}
sum = 0; mx = 0;
prev = n;
for (int i = n - 1; i >= ind; i--) {
if (c[i] >= mx) {
if (i < n - 1) {
ans += ((long long) (prev - i - 1) * mx - sum) % mod; ans %= mod;
}
mx = c[i];
prev = i;
sum = 0;
} else {
sum += c[i];
}
}
}
cout << ans << '\n';
}
