#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define re register
#define X first
#define Y second
#define mp std::make_pair
#define ohh(hhh...) fprintf(stderr,hhh)
typedef long long ll;
typedef unsigned long long ull;
typedef std::pair<int,int>pii;
const int MAXN=2e5+10;
int n,m;
ll a[MAXN],b[MAXN]; 
ll read()
{
	ll x=0,w=0;
	char ch=0;
	while(ch<'0'||ch>'9') w|=ch=='-',ch=getchar();
	while('0'<=ch&&ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return w?-x:x;
} 
ll gcd(ll a,ll b)
{
	if(!b) return a;
	return gcd(b,a%b); 
}
int main()
{
	n=read();
	m=read();
	for(re int i=1;i<=n;i++) a[i]=read();
	for(re int i=1;i<=m;i++) b[i]=read();
	std::sort(a+1,a+n+1);
	n=std::unique(a+1,a+n+1)-a-1;
	re ll now=a[2]-a[1];
	for(re int i=2;i<=n;i++) now=gcd(now,a[i]-a[1]);
	for(re int j=1;j<=m;j++)
	{
		if(n==1) printf("%lld ",a[1]+b[j]);
		else printf("%lld ",gcd(now,a[1]+b[j])); 
	} 
	return 0; 
}

### Prompt

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

You are given two positive integer sequences a_1, …, a_n and b_1, …, b_m. For each j = 1, …, m find the greatest common divisor of a_1 + b_j, …, a_n + b_j.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5).

The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^{18}).

The third line contains m integers b_1, …, b_m (1 ≤ b_j ≤ 10^{18}).

Output

Print m integers. The j-th of them should be equal to GCD(a_1 + b_j, …, a_n + b_j).

Example

Input


4 4
1 25 121 169
1 2 7 23


Output


2 3 8 24

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define re register
#define X first
#define Y second
#define mp std::make_pair
#define ohh(hhh...) fprintf(stderr,hhh)
typedef long long ll;
typedef unsigned long long ull;
typedef std::pair<int,int>pii;
const int MAXN=2e5+10;
int n,m;
ll a[MAXN],b[MAXN]; 
ll read()
{
	ll x=0,w=0;
	char ch=0;
	while(ch<'0'||ch>'9') w|=ch=='-',ch=getchar();
	while('0'<=ch&&ch<='9') x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return w?-x:x;
} 
ll gcd(ll a,ll b)
{
	if(!b) return a;
	return gcd(b,a%b); 
}
int main()
{
	n=read();
	m=read();
	for(re int i=1;i<=n;i++) a[i]=read();
	for(re int i=1;i<=m;i++) b[i]=read();
	std::sort(a+1,a+n+1);
	n=std::unique(a+1,a+n+1)-a-1;
	re ll now=a[2]-a[1];
	for(re int i=2;i<=n;i++) now=gcd(now,a[i]-a[1]);
	for(re int j=1;j<=m;j++)
	{
		if(n==1) printf("%lld ",a[1]+b[j]);
		else printf("%lld ",gcd(now,a[1]+b[j])); 
	} 
	return 0; 
}

```