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

#define FOR(x,n) for(int x=0;x<n;x++)
#define mp make_pair
#define PI 3.14159265358979323846264338327950288
typedef long long ll;
typedef pair<int,int> ii;

const int MX = 102;
const int MX2 = MX * MX;
const ll MOD = 1e9 + 7;

int n;
ll c[MX], b[MX], pref[MX2 + 1];
ll dp[MX][MX2];

ll solve(ll x) {
  ll curr = x;
  ll req = 0;

  dp[0][0] = 1;
  for(int i = 1; i <= n; i++) {
    req += curr;
    pref[0] = 0;
    for(int j = 1; j <= MX2; j++) {
      pref[j] = (pref[j - 1] + dp[i - 1][j - 1]) % MOD;
    }

    FOR(j,MX2) {
      if(j < req) {
	dp[i][j] = 0;
	continue;
      }
      int lo = max((ll)0, j - c[i - 1]);
      dp[i][j] = (pref[j + 1] - pref[lo] + MOD) % MOD;
    }
    curr += b[i - 1];
  }

  ll ans = 0;
  FOR(j,MX2) {
    ans = (ans + dp[n][j]) % MOD;
  }
  return ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  // mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

  memset(dp, 0, sizeof dp);

  cin >> n;
  FOR(i,n) cin >> c[i];
  FOR(i,n-1) cin >> b[i];

  ll y = 0;
  ll curr = 0;
  FOR(i,n-1) {
    curr += b[i];
    y += curr;
  }

  int xbig = (n * MX - y)/n + 1;
  int xsmall = xbig;
  while(n * xsmall + y >= 0) xsmall--;

  map<int,int> ans;
  for(int x = xsmall + 1; x < xbig; x++) {
    ans[x] = solve(x);
  }

  int anssmall = 1;
  FOR(i,n) {
    anssmall = (1ll * anssmall * (c[i] + 1)) % MOD;
  }

  int q;
  cin >> q;

  FOR(i,q) {
    int x;
    cin >> x;
    if(x <= xsmall) cout << anssmall << '\n';
    else if(x >= xbig) cout << "0\n";
    else cout << ans[x] << '\n';
  }
}

// UPDATE N WHEN UR DONE AAAAAAAAAAAAAAAAAAAAAAAAAAA (if it applies lol)

### 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>
using namespace std;

#define FOR(x,n) for(int x=0;x<n;x++)
#define mp make_pair
#define PI 3.14159265358979323846264338327950288
typedef long long ll;
typedef pair<int,int> ii;

const int MX = 102;
const int MX2 = MX * MX;
const ll MOD = 1e9 + 7;

int n;
ll c[MX], b[MX], pref[MX2 + 1];
ll dp[MX][MX2];

ll solve(ll x) {
  ll curr = x;
  ll req = 0;

  dp[0][0] = 1;
  for(int i = 1; i <= n; i++) {
    req += curr;
    pref[0] = 0;
    for(int j = 1; j <= MX2; j++) {
      pref[j] = (pref[j - 1] + dp[i - 1][j - 1]) % MOD;
    }

    FOR(j,MX2) {
      if(j < req) {
	dp[i][j] = 0;
	continue;
      }
      int lo = max((ll)0, j - c[i - 1]);
      dp[i][j] = (pref[j + 1] - pref[lo] + MOD) % MOD;
    }
    curr += b[i - 1];
  }

  ll ans = 0;
  FOR(j,MX2) {
    ans = (ans + dp[n][j]) % MOD;
  }
  return ans;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  // mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

  memset(dp, 0, sizeof dp);

  cin >> n;
  FOR(i,n) cin >> c[i];
  FOR(i,n-1) cin >> b[i];

  ll y = 0;
  ll curr = 0;
  FOR(i,n-1) {
    curr += b[i];
    y += curr;
  }

  int xbig = (n * MX - y)/n + 1;
  int xsmall = xbig;
  while(n * xsmall + y >= 0) xsmall--;

  map<int,int> ans;
  for(int x = xsmall + 1; x < xbig; x++) {
    ans[x] = solve(x);
  }

  int anssmall = 1;
  FOR(i,n) {
    anssmall = (1ll * anssmall * (c[i] + 1)) % MOD;
  }

  int q;
  cin >> q;

  FOR(i,q) {
    int x;
    cin >> x;
    if(x <= xsmall) cout << anssmall << '\n';
    else if(x >= xbig) cout << "0\n";
    else cout << ans[x] << '\n';
  }
}

// UPDATE N WHEN UR DONE AAAAAAAAAAAAAAAAAAAAAAAAAAA (if it applies lol)

```