#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];
}
vector<long long> p2(n + 1);
p2[0] = 1;
for (int i = 1; i <= n; i++) {
p2[i] = (p2[i - 1] * 2) % mod;
}
long long ans = 0;
for (int l = n - 1; l >= 0; l--) {
for (int r = l + 2; r < n; r++) {
for (int lv = 0; lv < 2; lv++) {
for (int rv = 0; rv < 2; rv++) {
int lval = (lv == 0 ? a[l] : b[l]);
int rval = (rv == 0 ? a[r] : b[r]);
int x = min(lval, rval);
long long ways = 1, in = 0;
for (int i = l + 1; i < r; i++) {
int new_in = 0;
if (a[i] < x) {
new_in = ((x - a[i]) * ways + in) % mod;
}
if (b[i] < x) {
new_in += ((x - b[i]) * ways + in) % mod; new_in %= mod;
}
in = new_in;
int tmp = 2;
if (a[i] >= x) tmp--;
if (b[i] >= x) tmp--;
ways *= tmp; ways %= mod;
}
long long lw = 1, rw = 1;
for (int i = 0; i < l; i++) {
int tmp = 2;
if (a[i] > lval) tmp--; if (b[i] > lval) tmp--;
lw *= tmp; lw %= mod;
}
for (int i = r + 1; i < n; i++) {
int tmp = 2;
if (a[i] > rval) tmp--; if (b[i] > rval) tmp--;
rw *= tmp; rw %= mod;
}
long long out = 1;
if (lval > rval) {
out = rw * p2[l] % mod;
} else if (rval > lval) {
out = lw * p2[n - 1 - r] % mod;
} else {
out = (p2[l + (n - 1 - r)] + mod -
(((p2[l] + mod - lw) % mod)
* ((p2[n - 1 - r] + mod - rw) % mod) % mod)) % mod;
}
ans += in * out % mod;
ans %= mod;
}
}
}
}
cout << ans << '\n';
}
