#include <bits/stdc++.h>
using namespace std;
// Use long long for integer types to prevent overflow, a good practice from the AC code.
#define int long long
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
const int inv2 = 500000004; // Correct modular inverse of 2
// Helper function to calculate sum from 1 to x, i.e., x*(x+1)/2
int sum_1_to_x(int x) {
x %= MOD;
int res = x * (x + 1) % MOD;
res = res * inv2 % MOD;
return res;
}
// Global arrays
int n;
int h[N], w[N], prefix[N], dp[N];
int l[N];
void Solve() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> h[i];
// Use true prefix sums, it's safer and simpler.
// long long is large enough.
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++) {
// Standard monotonic stack to find the first element to the left shorter than h[i]
while (!st.empty() && h[st.top()] >= h[i]) {
st.pop();
}
l[i] = st.empty() ? 0 : st.top();
st.push(i);
// --- Calculate Contribution to Final Answer ---
// 1. Contribution from rectangles fully inside the current bar (h[i], w[i])
ans = (ans + sum_1_to_x(h[i]) * sum_1_to_x(w[i])) % MOD;
// 2. Contribution from rectangles extending from the left.
// This is based on the state of the previous shorter bar, dp[l[i]],
// plus the contribution of the block between l[i] and i.
// Width of the block from l[i]+1 to i-1
int W_block_before = (prefix[i - 1] - prefix[l[i]]);
W_block_before = (W_block_before % MOD + MOD) % MOD;
// Temporary sum representing all valid rectangles ending before i that can be extended by w[i]
int temp_sum = (dp[l[i]] + W_block_before * sum_1_to_x(h[i])) % MOD;
ans = (ans + temp_sum * w[i]) % MOD;
// --- Update DP state for the next iteration ---
// Total width of the new continuous block with minimum height h[i]
int W_total_block = (prefix[i] - prefix[l[i]]);
W_total_block = (W_total_block % MOD + MOD) % MOD;
// The new DP state is the state from the previous shorter bar plus the
// contribution of the new block.
dp[i] = (dp[l[i]] + W_total_block * sum_1_to_x(h[i])) % MOD;
}
cout << ans << endl;
}
signed main() {
// Fast I/O
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
Solve();
return 0;
}
