The problem models a sequential arrangement where each position $i$ offers two distinct categories of elements. Category X occupies exactly one slot, while Category Y spans two consecutive slots $(i-1, i)$. The objective is to compute the total number of valid configurations modulo $10^9+7$.

A two-state dynamic programming approach efficiently captures the positional dependencies. Let $dp[i][0]$ denote the number of valid sequences completing at index $i$ by placing an element from Category X. Let $dp[i][1]$ represent sequences finishing at $i$ using a element from Category Y, wich necessarily overlaps with index $i-1$.

Initialization At the first index, base counts map directly to input arrays: $$dp[1][0] = \text{count}_X[1]$$ $$dp[1][1] = \text{count}_Y[1]$$ The final result aggregates both terminal states: $$\text{answer} = dp[n][0] + dp[n][1]$$

Transition Dynamics Moving from index $i-1$ to $i$ requires evaluating how prior placements constrain curent choices.

When computing $dp[i][0]$, the preceding state determines the multiplier:

• If index $i-1$ concluded with Category X, all available items at $i$ ($\text{count}_X[i]$) combine with non-overlapping Category Y options from the previous tier ($\text{count}_Y[i-1]$). Contribution: $$dp[i-1][0] \times (\text{count}_X[i] + \text{count}_Y[i-1])$$
• If index $i-1$ concluded with Category Y, one placement slot is structurally tied to the overlap. We must subtract one from the previous tier's Category Y pool to avoid double-counting, clamped at zero. Contribution: $$dp[i-1][1] \times (\text{count}_X[i] + \max(\text{count}_Y[i-1] - 1, 0))$$

For $dp[i][1]$, the recurrence simplifies since placing a Category Y item at $i$ is independent of the specific type used at $i-1$, provided the sequence was valid. Each valid prefix accumulates $\text{count}_Y[i]$ new possibilities: $$dp[i][1] = (dp[i-1][0] + dp[i-1][1]) \times \text{count}_Y[i]$$

Applying these transitions iteratively yields a linear time complexity of $O(n)$.

Space Optimization Since each step $i$ depends exclusively on values from $i-1$, the full table is unnecessary. Maintaining only four scalar variables eliminates $O(n)$ auxiliary memory, reducing space complexity to $O(1)$.

Implementation Below is a production-ready C++ solution incorporating modular arithmetic safeguards and the optimized scalar state tracking.

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
using ll = long long;

constexpr int MOD = 1e9 + 7;

void solve() {
    int n;
    if (!(cin >> n)) return;

    vector<ll> x_items(n + 1), y_items(n + 1);
    for (int i = 1; i <= n; ++i) cin >> x_items[i];
    for (int i = 1; i < n; ++i) cin >> y_items[i];

    // Current and previous state scalars
    ll prev_single = x_items[1];
    ll prev_dual   = y_items[1];

    for (int i = 2; i <= n; ++i) {
        ll curr_single = (prev_single * (x_items[i] + y_items[i - 1]) % MOD +
                          prev_dual * (x_items[i] + max(y_items[i - 1] - 1, 0LL)) % MOD) % MOD;
        
        ll curr_dual = (prev_single + prev_dual) % MOD * y_items[i] % MOD;

        prev_single = curr_single;
        prev_dual   = curr_dual;
    }

    cout << (prev_single + prev_dual) % MOD << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}