#include <bits/stdc++.h>
using namespace std;
const int md=1000000007,MID=111*111,ST=100000;
int n,q,x,le,ri,i,ii,i1,j,lst,len,lensum,cur,r,b[111],c[111];
long long f[2][MID];
map<int,int> a;
int main() {
  scanf("%d",&n);
  for (i=1; i<=n; i++) scanf("%d",&c[i]);
  for (i=1; i<n; i++) scanf("%d",&b[i]);
  f[0][0]=1;
  for (i=1; i<=n; i++) {
    ii=i&1; i1=1-ii;
    cur+=c[i];
    a[100000]=-1;
    for (j=0; j<=cur; j++) {
      le=0; ri=2*ST;
      while (le<ri) {
        x=(le+ri)/2+1;
        if ((x-ST)*i+lensum<=j) le=x; else ri=x-1;
      }
      a[ri-ST]=-1;
      f[ii][j]=(f[ii][j]+f[i1][j-lst])%md;
    }
    len+=b[i];
    lensum+=len;
  }
  scanf("%d",&q);
  while (q--) {
    scanf("%d",&x);
    if (x>c[1]) { puts("0"); continue; }
    auto it=a.lower_bound(x);
    if (it->second>=0) { printf("%d\n",it->second); continue; }
    x=it->first;
    r=cur=len=lensum=0;
    memset(f,0,sizeof(f));
    f[0][0]=1;
    for (i=1; i<=n; i++) {
      ii=i&1; i1=1-ii;
      cur+=c[i];
      for (j=1; j<=cur; j++) if ((f[i1][j]+=f[i1][j-1])>=md) f[i1][j]-=md;
      le=x*i+lensum;
      if (le>=MID) break;
      for (j=0; j<le; j++) f[ii][j]=0;
      for (; j<=cur; j++) {
        f[ii][j]=f[i1][j];
        if (j>c[i] && (f[ii][j]-=f[i1][j-c[i]-1])<0) f[ii][j]+=md;
      }
      len+=b[i];
      lensum+=len;
    }
    if (i>n) for (j=0; j<=cur; j++) r=(r+f[n&1][j])%md;
    printf("%d\n",r);
    a[x]=r;
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>
using namespace std;
const int md=1000000007,MID=111*111,ST=100000;
int n,q,x,le,ri,i,ii,i1,j,lst,len,lensum,cur,r,b[111],c[111];
long long f[2][MID];
map<int,int> a;
int main() {
  scanf("%d",&n);
  for (i=1; i<=n; i++) scanf("%d",&c[i]);
  for (i=1; i<n; i++) scanf("%d",&b[i]);
  f[0][0]=1;
  for (i=1; i<=n; i++) {
    ii=i&1; i1=1-ii;
    cur+=c[i];
    a[100000]=-1;
    for (j=0; j<=cur; j++) {
      le=0; ri=2*ST;
      while (le<ri) {
        x=(le+ri)/2+1;
        if ((x-ST)*i+lensum<=j) le=x; else ri=x-1;
      }
      a[ri-ST]=-1;
      f[ii][j]=(f[ii][j]+f[i1][j-lst])%md;
    }
    len+=b[i];
    lensum+=len;
  }
  scanf("%d",&q);
  while (q--) {
    scanf("%d",&x);
    if (x>c[1]) { puts("0"); continue; }
    auto it=a.lower_bound(x);
    if (it->second>=0) { printf("%d\n",it->second); continue; }
    x=it->first;
    r=cur=len=lensum=0;
    memset(f,0,sizeof(f));
    f[0][0]=1;
    for (i=1; i<=n; i++) {
      ii=i&1; i1=1-ii;
      cur+=c[i];
      for (j=1; j<=cur; j++) if ((f[i1][j]+=f[i1][j-1])>=md) f[i1][j]-=md;
      le=x*i+lensum;
      if (le>=MID) break;
      for (j=0; j<le; j++) f[ii][j]=0;
      for (; j<=cur; j++) {
        f[ii][j]=f[i1][j];
        if (j>c[i] && (f[ii][j]-=f[i1][j-c[i]-1])<0) f[ii][j]+=md;
      }
      len+=b[i];
      lensum+=len;
    }
    if (i>n) for (j=0; j<=cur; j++) r=(r+f[n&1][j])%md;
    printf("%d\n",r);
    a[x]=r;
  }
  return 0;
}

```