#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define fors(i, a, b) for(int i = (a); i < (b); ++i)
#define all(vec) vec.begin(),vec.end()
using namespace std;
const int maxn = 105;
int n, q;
const int mod = 1e9 + 7;
int dp[maxn][maxn*maxn];
int s[maxn][maxn*maxn];
int b[maxn], c[maxn];
int sol(int x){
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    fors(i,0,maxn*maxn) s[0][i] = 1;
    int low = 0, up = 0;
    for(int i = 1; i <= n; ++i){
        up += c[i];
        low += b[i-1]+x;
        for(int j = max(0, low); j <= up; ++j){
            int l = max(0, j-c[i]), r = j;
            if(l == 0) dp[i][j] = s[i-1][r];
            else dp[i][j] = (s[i-1][r]-s[i-1][l-1])%mod;
        }
        s[i][0] = dp[i][0];
        fors(j,1,maxn*maxn) s[i][j] = (dp[i][j]+s[i][j-1])%mod;
    }
   return (s[n][maxn*maxn-1]+mod)%mod;
}
const int sed = 10007;
map<int,int> mp;
int main()
{
    cin>>n; fors(i,1,n+1) cin>>c[i];
    fors(i,1,n) cin>>b[i]; cin>>q;
    fors(i,1,n) b[i]+=b[i-1];
    //int x; cin>>x;
    int l = 0x3f3f3f3f;
    int r = 0x3f3f3f3f;

    int low = 0, up = 0;
    fors(i,1,n+1){
        up += c[i];
        low += b[i-1];//low + i*x <= up
        r = min(r, (up-low)/i);
        l = min(l, -low/i);
    }
    //if( n > 10) cout<<"l:"<<l<<" r:"<<r<<endl;
    l--; r++;
    fors(i,l,r+1) mp[i] = sol(i);
    //if(n > 10) cout<<"sol:"<<sol(l)<<endl;
    int ca = 0;
    while(q--){
        int x; scanf("%d", &x);
        if(x < l) x = l;
        if(x > r) printf("0\n");
        else printf("%d\n", mp[x]);
    }

	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

This is the hard version of the problem. The only difference is that in this version 1 ≤ q ≤ 10^5. 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 (1 ≤ q ≤ 10^5).

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
5
-1 0 1 -100000 100000


Output


56
28
4
60
0

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 ll long long
#define pb push_back
#define lowbit(x) ((x)&(-(x)))
#define mid ((l+r)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define fors(i, a, b) for(int i = (a); i < (b); ++i)
#define all(vec) vec.begin(),vec.end()
using namespace std;
const int maxn = 105;
int n, q;
const int mod = 1e9 + 7;
int dp[maxn][maxn*maxn];
int s[maxn][maxn*maxn];
int b[maxn], c[maxn];
int sol(int x){
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    fors(i,0,maxn*maxn) s[0][i] = 1;
    int low = 0, up = 0;
    for(int i = 1; i <= n; ++i){
        up += c[i];
        low += b[i-1]+x;
        for(int j = max(0, low); j <= up; ++j){
            int l = max(0, j-c[i]), r = j;
            if(l == 0) dp[i][j] = s[i-1][r];
            else dp[i][j] = (s[i-1][r]-s[i-1][l-1])%mod;
        }
        s[i][0] = dp[i][0];
        fors(j,1,maxn*maxn) s[i][j] = (dp[i][j]+s[i][j-1])%mod;
    }
   return (s[n][maxn*maxn-1]+mod)%mod;
}
const int sed = 10007;
map<int,int> mp;
int main()
{
    cin>>n; fors(i,1,n+1) cin>>c[i];
    fors(i,1,n) cin>>b[i]; cin>>q;
    fors(i,1,n) b[i]+=b[i-1];
    //int x; cin>>x;
    int l = 0x3f3f3f3f;
    int r = 0x3f3f3f3f;

    int low = 0, up = 0;
    fors(i,1,n+1){
        up += c[i];
        low += b[i-1];//low + i*x <= up
        r = min(r, (up-low)/i);
        l = min(l, -low/i);
    }
    //if( n > 10) cout<<"l:"<<l<<" r:"<<r<<endl;
    l--; r++;
    fors(i,l,r+1) mp[i] = sol(i);
    //if(n > 10) cout<<"sol:"<<sol(l)<<endl;
    int ca = 0;
    while(q--){
        int x; scanf("%d", &x);
        if(x < l) x = l;
        if(x > r) printf("0\n");
        else printf("%d\n", mp[x]);
    }

	return 0;
}

```