#include <bits/stdc++.h>

#define all(a) begin(a), end(a)
#define forn(i, n) for (int i = 0; i < n; ++i)

using namespace std;
using ll = long long;

const int MOD = 1e9 + 7;
const int N = 103;
const int SUM = 101 * 101;

int dp[N][SUM];
int n, a[N], b[N];
int pb[N];

void solve() {
    // for all i sum - ((i-1)*b_1 + .. +b_{i-1}) >= x * i
    cin >> n;

    forn(i, n) {
        cin >> a[i];
    }

    pb[0] = 0;
    forn(i, n - 1) {
        cin >> b[i];
        pb[i + 1] = pb[i] + b[i];
    }

    for (int i = 1; i < n; ++i) {
        pb[i] += pb[i - 1];
    }

    int x;
    cin >> x >> x;

    forn(i, a[0] + 1) {
        dp[0][i] = 1;
    }

    forn(i, n - 1) {
        ll val = 1ll * (i + 1) * x + pb[i];
        val = max(val, 0ll);
        for (int j = val; j < SUM; ++j) {
            if (dp[i][j] == 0) continue;
            for (int nxt = 0; nxt <= a[i + 1]; ++nxt) {
                dp[i + 1][j + nxt] = (dp[i + 1][j + nxt] + dp[i][j]) % MOD;
            }
        }
    }

    int res = 0;
    ll lower = 1ll * n * x + pb[n - 1];
    lower = max(lower, 0ll);

    for (int i = lower; i < SUM; ++i) res = (res + dp[n - 1][i]) % MOD;
    cout << res << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    solve();

    return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.

There is a process that takes place on arrays a and b of length n and length n-1 respectively. 

The process is an infinite sequence of operations. Each operation is as follows: 

  * First, choose a random integer i (1 ≤ i ≤ n-1). 
  * Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer). 

See notes for an example of an operation.

It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.

You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.

Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.

Input

The first line contains a single integer n (2 ≤ n ≤ 100).

The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).

The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).

The fourth line contains a single integer q (q=1).

The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).

Output

Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.

Example

Input


3
2 3 4
2 1
1
-1


Output


56

Note

The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).

Examples of arrays a that are not good: 

  * a = [3, 2, 3] is not good because a_1 > c_1; 
  * a = [0, -1, 3] is not good because a_2 < 0. 



One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.

Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2.

### Response

```cpp
#include <bits/stdc++.h>

#define all(a) begin(a), end(a)
#define forn(i, n) for (int i = 0; i < n; ++i)

using namespace std;
using ll = long long;

const int MOD = 1e9 + 7;
const int N = 103;
const int SUM = 101 * 101;

int dp[N][SUM];
int n, a[N], b[N];
int pb[N];

void solve() {
    // for all i sum - ((i-1)*b_1 + .. +b_{i-1}) >= x * i
    cin >> n;

    forn(i, n) {
        cin >> a[i];
    }

    pb[0] = 0;
    forn(i, n - 1) {
        cin >> b[i];
        pb[i + 1] = pb[i] + b[i];
    }

    for (int i = 1; i < n; ++i) {
        pb[i] += pb[i - 1];
    }

    int x;
    cin >> x >> x;

    forn(i, a[0] + 1) {
        dp[0][i] = 1;
    }

    forn(i, n - 1) {
        ll val = 1ll * (i + 1) * x + pb[i];
        val = max(val, 0ll);
        for (int j = val; j < SUM; ++j) {
            if (dp[i][j] == 0) continue;
            for (int nxt = 0; nxt <= a[i + 1]; ++nxt) {
                dp[i + 1][j + nxt] = (dp[i + 1][j + nxt] + dp[i][j]) % MOD;
            }
        }
    }

    int res = 0;
    ll lower = 1ll * n * x + pb[n - 1];
    lower = max(lower, 0ll);

    for (int i = lower; i < SUM; ++i) res = (res + dp[n - 1][i]) % MOD;
    cout << res << "\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    solve();

    return 0;
}
```