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.
/*~Rainybunny~*/

#include <cstdio>
#include <cstring>

#define rep( i, l, r ) for ( int i = l, rep##i = r; i <= rep##i; ++i )
#define per( i, r, l ) for ( int i = r, per##i = l; i >= per##i; --i )


const int MAXN = 101, MOD = 1e9 + 7;
int n, c[MAXN + 5], b[MAXN + 5];
int sc[MAXN + 5], sb[MAXN + 5], sb2[MAXN + 5];
int h[MAXN + 5], ans[MAXN + 5];

inline void chkmax( int& a, const int b ) { a < b && ( a = b ); }
inline int imin( const int a, const int b ) { return a < b ? a : b; }
inline int imax( const int a, const int b ) { return a < b ? b : a; }
inline int sub( int a, const int b ) { return ( a -= b ) < 0 ? a + MOD : a; }
inline int add( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }

inline int solve( const int lim ) {
	static int f[MAXN * MAXN + 5], g[MAXN * MAXN + 5];
	memset( f, 0, sizeof f ), f[0] = 1;
	rep ( i, 1, n ) {
		g[0] = f[0];
		rep ( j, 1, sc[i] ) g[j] = add( g[j - 1], f[j] );
		rep ( j, 0, lim * i + sb[i - 1] - 1 ) f[j] = 0;
		rep ( j, imax( 0, lim * i + sb[i - 1] ), sc[i] ) {
			f[j] = sub( g[j], j > c[i] ? g[j - c[i] - 1] : 0 );
		}
	}
	int ret = 0;
	rep ( i, imax( 0, lim * n + sb[n - 1] ), sc[n] ) ret = add( ret, f[i] );
	return ret;
}

int main() {
	scanf( "%d", &n );
	rep ( i, 1, n ) scanf( "%d", &c[i] ), sc[i] = sc[i - 1] + c[i];
	rep ( i, 1, n - 1 ) scanf( "%d", &b[i] );
	
	int mx = 0xcfcfcfcf;
	rep ( i, 1, n ) sb[i] = sb[i - 1] + b[i];
	rep ( i, 1, n ) sb[i] += sb[i - 1];
	rep ( i, 1, n ) chkmax( mx, ( sb[i - 1] + i - 1 ) / i );
	
	rep ( i, 0, MAXN ) ans[i] = solve( i - mx );
	
	int q, x; scanf( "%d", &q );
	while ( q-- ) {
		scanf( "%d", &x ), x = imin( MAXN - mx, imax( x, -mx ) );
		printf( "%d\n", ans[x + mx] );
	}
	return 0;
}

