/*
Assume we already know the number of rectangles on the first (i - 1) fence.
Define f(i) as (i * i + i)/2;
If we are to add the i-th fence, then the new rectangles can be categorized into 3:
* It lies completely on the i-th fence. There are f(h[i]) * f(w[i]) such rectangles.
* The right end ends on the i-th fence while the left end lies on taller fence to the left of it.
We don't want to consider the taller fence that comes before a fence lower than the i-th fence.
Let's say the first lower fence to the left of the i-th fence is the j-th fence.
There are (sumWidth(j + 1, i) * w[i]) * f(h[i]) such rectangles.
* The right end ends on the i-th fence while the left end lies on lower fences.
Assume we keep the lower fences on a stack:
h_i1 < h_i2 < h_i3 ...
We can create rectangles whose height is at most h_i3 using the (i2 + 1)-th fence upto the i3-th fence.
There are (sumWidth(i2 + 1, i3) * w[i]) * f(h_i3) such rectangles.
We can do the same calculation for the (i1 + 1)-th fence upto the i2-th fence, and so on.
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long int
#define MOD 1000000007
#define MAXN 100000
int n;
LL h[MAXN + 3], w[MAXN + 3], sumWidth[MAXN + 3], sumFhw = 0, ans = 0;
stack<int> st;
LL f(LL k) {
LL k1 = k + 1;
return k%2 == 0 ? ((k/2)%MOD * (k1)%MOD)%MOD : (k%MOD * (k1/2)%MOD)%MOD;
}
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> h[i];
}
sumWidth[0] = 0;
for (int i = 1; i <= n; i++) {
cin >> w[i];
sumWidth[i] = sumWidth[i - 1] + w[i];
}
for (int i = 1; i <= n; i++) {
LL rectanglesOnTheCurrentFence = (f(h[i]) * f(w[i]))%MOD;
ans = (ans + rectanglesOnTheCurrentFence)%MOD;
while (!st.empty() && h[st.top()] >= h[i]) {
int top = st.top(); st.pop();
LL totalWidth = sumWidth[top] - (st.empty() ? 0 : sumWidth[st.top()]);
LL fhw = (f(h[top]) * (totalWidth%MOD))%MOD;
sumFhw = (sumFhw - fhw + MOD)%MOD;
}
int firstLowerIdx = (st.empty() ? 0 : st.top());
LL rectanglesOnTheCurrentFenceAndTallerNeighbors = (((sumWidth[i - 1] - sumWidth[firstLowerIdx])%MOD * w[i])%MOD * f(h[i]))%MOD;
ans = (ans + rectanglesOnTheCurrentFenceAndTallerNeighbors)%MOD;
LL rectanglesOnTheCurrentFenceAndLowerNeighbors = (sumFhw * w[i])%MOD;
ans = (ans + rectanglesOnTheCurrentFenceAndLowerNeighbors)%MOD;
LL totalWidth = sumWidth[i] - (st.empty() ? 0 : sumWidth[st.top()]);
sumFhw = (sumFhw + (f(h[i]) * (totalWidth%MOD))%MOD)%MOD;
st.push(i);
}
cout << ans << endl;
}
