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;
typedef long long ll;
const int maxn=1e2+50;//注意修改大小
const int maxm=1e4+50;
const int MAXM=1e5+50;
const ll Mod=1e9+7;
long long read(){long long x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}while(isdigit(c)){x=x*10+c-'0';c=getchar();}return x*f;}
ll qpow(ll x,ll q,ll Mod){ll ans=1;while(q){if(q&1)ans=ans*x%Mod;q>>=1;x=(x*x)%Mod;}return ans%Mod;}
ll n,q,a[maxn],b[maxn],c[maxn],api[maxn],bpi[maxn],cpi[maxn],bppi[maxn],ans[2*MAXM],dp[2][MAXM];
int main(){
    // ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>c[i];
    for(int i=1;i<=n-1;i++)cin>>b[i];
    for(int i=1;i<=n-1;i++)bpi[i]=bpi[i-1]+b[i];
    for(int i=1;i<=n-1;i++)bppi[i]=bppi[i-1]+bpi[i];
    for(int i=1;i<=n;i++)cpi[i]=cpi[i-1]+c[i];
    // for(int i=1;i<=n;i++)cout<<cpi[i]<<' ';cout<<endl;
    // for(int i=1;i<n;i++)cout<<cpi[i]<<' ';cout<<endl;
    // for(int i=1;i<n;i++)cout<<bppi[i]<<' ';cout<<endl;
    cin>>q;
    for(int i=0;i<=2e5;i++)ans[i]=-1;
    ll up=-1e18,down=1e18,maxxc=0;
    for(int i=1;i<=n;i++){
        up=max(up,(ll)ceil(-bppi[i-1]*1.0/i)),down=min(down,(ll)floor(-bppi[i-1]*1.0/i));
        maxxc=max(maxxc,c[i]);
    }
    up+=maxxc;
    while(q--){
        int x;cin>>x;
        if(ans[x+MAXM]!=-1){
            cout<<ans[x+MAXM]<<endl;
            continue;
        }
        if(x<down){
            ll ans=1;
            for(int i=1;i<=n;i++)ans=ans*(c[i]+1)%Mod;
            cout<<ans<<endl;
            continue;
        }
        if(x>up){
            cout<<0<<endl;
            continue;
        }
        memset(dp,0,sizeof(dp));
        ll now=1,nex=0;
        for(int i=0;i<=c[1];i++)dp[1][i]=1;
        for(int i=1;i<=n;i++){
            memset(dp[nex],0,sizeof(dp[nex]));
            for(int j=max(0ll,x*i+bppi[i-1]);j<=cpi[i];j++){
                dp[nex][j]=(dp[nex][j]+dp[now][min(cpi[i-1],(ll)j)]+Mod)%Mod;
                if(j-c[i]-1>=0){
                    dp[nex][j]=(dp[nex][j]-dp[now][j-c[i]-1]+Mod)%Mod;
                }
            }
            for(int j=1;j<=cpi[i];j++)dp[nex][j]=(dp[nex][j]+dp[nex][j-1])%Mod;
            swap(nex,now);
        }
        ans[x+MAXM]=dp[now][cpi[n]];
        cout<<ans[x+MAXM]<<endl;
    }
    return 0;
}