#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+ 5;
const long long inv2 = 500000004; // Correct
const int MOD = 1e9 + 7;
long long h[N], w[N], prefix[N], dp[N];
int n;
int l[N];
void Solve() {
cin >> n;
for(int i = 1; i <= n; i++) cin >> h[i];
// Revert to storing true prefix sums
for(int i = 1; i <= n; i++) {
cin >> w[i];
prefix[i] = prefix[i - 1] + w[i];
}
long long ans = 0;
stack<int> st;
for(int i = 1; i <= n; i++) {
while(!st.empty() && h[st.top()] >= h[i]) st.pop();
if(st.empty()) l[i] = 0;
else l[i] = st.top();
st.push(i);
// Calculate W from true sums. This is simpler and safer
// since (prefix[i-1] - prefix[l[i]]) will be non-negative if all w[i] are >= 0.
// To be fully robust for negative widths, you could do:
long long W = (prefix[i - 1] - prefix[l[i]]);
W = (W % MOD + MOD) % MOD;
// The rest of your logic remains the same
ans = (ans + h[i] * (h[i] + 1) % MOD * inv2 % MOD * w[i] % MOD * (w[i] + 1) % MOD * inv2 % MOD) % MOD;
ans = (ans + h[i] * (h[i] + 1) % MOD * inv2 % MOD * w[i] % MOD * W % MOD) % MOD;
long long current_dp_term = h[i] * (h[i] + 1) % MOD * inv2 % MOD * (W + (w[i] % MOD) + MOD) % MOD;
dp[i] = (current_dp_term + dp[l[i]]) % MOD;
ans = (ans + w[i] % MOD * h[l[i]] % MOD * dp[l[i]] % MOD + MOD) % MOD;
}
cout << ans << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL); cout.tie(NULL);
Solve();
}
