Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long
constexpr int MAXN = 100;
constexpr int MOD = 1000000007;
constexpr int MAXTOLERANCE = 10000001; //10001; // really? TODO
int c[MAXN], b[MAXN], magic[MAXN];
int dp1[MAXTOLERANCE], dp2[MAXTOLERANCE];
int coalesce[MAXTOLERANCE+1];
int* const coal=coalesce+MAXTOLERANCE/2;
int32_t main(){
    ios_base::sync_with_stdio(false);
    int n,q;
    cin>>n;
    for (int i=0;i<n;++i){
        cin>>c[i];
    }
    for(int i=1;i<n;++i){
        cin>>b[i];
    }
    cin>>q;
    for(int j=0;j<q;++j){
        int x;
        cin>>x;
        /*if (x>MAXN){
            cout<<0<<'\n';
            continue;
        }*/
        magic[0]=x;
        for(int i=1;i<n;++i){
            magic[i]=magic[i-1]+b[i];
        }
        int* prev=dp1+MAXTOLERANCE/2;
        //fill_n(prev-MAXTOLERANCE/2, MAXTOLERANCE, 0);
        prev[x]=1;
        int prev_low=x;
        int prev_high=x;
        int* curr=dp2+MAXTOLERANCE/2;
        //fill_n(curr-MAXTOLERANCE/2, MAXTOLERANCE, 0);
        for(int i=0;i<n-1;++i){
            //fill_n(curr-MAXTOLERANCE/2, MAXTOLERANCE, 0);
            //fill_n(coal-MAXTOLERANCE/2, MAXTOLERANCE+1, 0);
            // TODO: optimize
            int coal_low=LLONG_MAX;
            int coal_high=LLONG_MIN;
            for(int k=prev_low;k<=prev_high;++k){
                int start = max(k, 0ll);
                if (start <= c[i] && prev[k]>0){
                    int c_start = min(max(magic[i+1]-(c[i]-k), -MAXTOLERANCE/2), MAXTOLERANCE/2+1);
                    int c_end = min(max(magic[i+1]-(start-k)+1, -MAXTOLERANCE/2), MAXTOLERANCE/2+1);
                    coal[c_start]+=prev[k];
                    coal[c_start]%=MOD;
                    coal[c_end]+=(MOD-prev[k]);
                    coal[c_end]%=MOD;
                    coal_low=min(coal_low,c_start);
                    coal_high=max(coal_high,c_end-1);
                }

                /*for(int kk=start;kk<=c[i];++kk){
                    int ind = magic[i+1]-(kk-k);
                    if (ind >= -MAXTOLERANCE/2 && ind <= MAXTOLERANCE/2){
                        curr[ind]+=prev[k];
                        curr[ind]%=MOD;
                    }
                }*/
            }
            int cum=0;
            for(int k=coal_low;k<=coal_high;++k){
                cum+=coal[k];
                cum%=MOD;
                curr[k]=cum;
            }
            fill(coal+coal_low,coal+coal_high+2,0);
            fill(prev+prev_low, prev+prev_high+1,0);
            prev_low=coal_low;
            prev_high=coal_high;
            swap(prev, curr);
            /*for(int k=-10;k<=10;++k){
                cout<<' '<<prev[k];
            }
            cout<<endl;
            cout<< prev_low<<' '<<prev_high<<endl;*/
        }
        int ans=0;
        {
            //fill_n(curr-MAXTOLERANCE/2, MAXTOLERANCE, 0);
            // TODO: optimize
            for(int k=prev_low;k<=prev_high;++k){
                int start = max(k, 0ll);
                for(int kk=start;kk<=c[n-1];++kk){
                    ans+=prev[k];
                    ans%=MOD;
                }
            }
            fill(prev+prev_low,prev+prev_high+1,0);
        }
        cout<<ans<<'\n';
    }
}