#include<bits/stdc++.h>
using namespace std;

int read();
#define LL long long
#define fr(i,l,r) for(int i = (l);i <= (r);++i)
#define rf(i,l,r) for(int i = (l);i >= (r);--i)
#define fo(i,l,r) for(int i = (l);i < (r);++i)
#define foredge(i,u,v) for(int i = fir[u],v;v = to[i],i;i = nxt[i])
#define ci const int &
#define cl const LL &
#define I inline void
#define filein(File) freopen(File".in","r",stdin)
#define fileout(File) freopen(File".out","w",stdout)

const int N = 105,M = 105,Q = 1e5 + 5,MOD = 1e9 + 7;
int c[N],b[N],dp[N][N*M],sumb[N],sum[N][N*M];
int n,m,qq;
pair<int,int> q[Q];
int ans[Q];
#define fi first
#define se second
void chkmod(int &x) {x += x >> 31 & MOD;}
void solve(int X) {
	memset(dp,0,sizeof dp);
	dp[0][0] = 1;
	fr(i,0,n) {
		if(i) fr(j,max(i*X+sumb[i-1],0),i*m) {
			if(j > c[i]) chkmod(dp[i][j] = sum[i-1][j] - sum[i-1][j-c[i]-1]);
			else dp[i][j] = sum[i-1][j];
		}
		sum[i][0] = dp[i][0];
		if(i != n) fr(j,1,(i+1)*m) chkmod(sum[i][j] = sum[i][j-1] + dp[i][j] - MOD);
		else fr(j,1,n*m) chkmod(sum[i][j] = sum[i][j-1] + dp[i][j] - MOD);
	}
}
int main() {
	//filein("c2");
	//fileout("c2");
	m = 100;
	cin >> n;
	fr(i,1,n) cin >> c[i];
	fo(i,1,n) {
		cin >> b[i];
		sumb[i] = sumb[i-1] + b[i];
	}
	fo(i,1,n) sumb[i] += sumb[i-1];
	int L = -sumb[n-1] / n - 1,R = (m * n - sumb[n-1]) / n + 1;
	cin >> qq;
	fr(i,1,qq) {
		cin >> q[i].fi,q[i].se = i;
		q[i].fi = min(q[i].fi,R);
		q[i].fi = max(q[i].fi,L);
	}
	sort(q+1,q+qq+1);
	fr(i,1,qq) {
		if(i == 1 || q[i].fi != q[i-1].fi)
			solve(q[i].fi);
		ans[q[i].se] = sum[n][n*m];
	}
	fr(i,1,qq) printf("%d\n",(ans[i] + MOD) % MOD);
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

int read();
#define LL long long
#define fr(i,l,r) for(int i = (l);i <= (r);++i)
#define rf(i,l,r) for(int i = (l);i >= (r);--i)
#define fo(i,l,r) for(int i = (l);i < (r);++i)
#define foredge(i,u,v) for(int i = fir[u],v;v = to[i],i;i = nxt[i])
#define ci const int &
#define cl const LL &
#define I inline void
#define filein(File) freopen(File".in","r",stdin)
#define fileout(File) freopen(File".out","w",stdout)

const int N = 105,M = 105,Q = 1e5 + 5,MOD = 1e9 + 7;
int c[N],b[N],dp[N][N*M],sumb[N],sum[N][N*M];
int n,m,qq;
pair<int,int> q[Q];
int ans[Q];
#define fi first
#define se second
void chkmod(int &x) {x += x >> 31 & MOD;}
void solve(int X) {
	memset(dp,0,sizeof dp);
	dp[0][0] = 1;
	fr(i,0,n) {
		if(i) fr(j,max(i*X+sumb[i-1],0),i*m) {
			if(j > c[i]) chkmod(dp[i][j] = sum[i-1][j] - sum[i-1][j-c[i]-1]);
			else dp[i][j] = sum[i-1][j];
		}
		sum[i][0] = dp[i][0];
		if(i != n) fr(j,1,(i+1)*m) chkmod(sum[i][j] = sum[i][j-1] + dp[i][j] - MOD);
		else fr(j,1,n*m) chkmod(sum[i][j] = sum[i][j-1] + dp[i][j] - MOD);
	}
}
int main() {
	//filein("c2");
	//fileout("c2");
	m = 100;
	cin >> n;
	fr(i,1,n) cin >> c[i];
	fo(i,1,n) {
		cin >> b[i];
		sumb[i] = sumb[i-1] + b[i];
	}
	fo(i,1,n) sumb[i] += sumb[i-1];
	int L = -sumb[n-1] / n - 1,R = (m * n - sumb[n-1]) / n + 1;
	cin >> qq;
	fr(i,1,qq) {
		cin >> q[i].fi,q[i].se = i;
		q[i].fi = min(q[i].fi,R);
		q[i].fi = max(q[i].fi,L);
	}
	sort(q+1,q+qq+1);
	fr(i,1,qq) {
		if(i == 1 || q[i].fi != q[i-1].fi)
			solve(q[i].fi);
		ans[q[i].se] = sum[n][n*m];
	}
	fr(i,1,qq) printf("%d\n",(ans[i] + MOD) % MOD);
	return 0;
}

```