#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define in inline
#define ld long db
namespace fast_io
{
	char buf[1<<12],*p1=buf,*p2=buf,sr[1<<23],z[23],nc;int C=-1,Z=0,Bi=0,ny;
	in char gc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<12,stdin),p1==p2)?EOF:*p1++;}
	in ll read()
	{
		ll x=0;ny=1;while(nc=gc(),(nc<48||nc>57)&&nc!=EOF)if(nc==45)ny=-1;Bi=1;if(nc<0)return nc;
		x=nc-48;while(nc=gc(),47<nc&&nc<58&&nc!=EOF)x=(x<<3)+(x<<1)+(nc^48),Bi++;return x*ny;
	}
	in db gf() {int a=read(),y=ny,b=(nc!='.')?0:read();return (b?a+(db)b/pow(10,Bi)*y:a);}
	in int gs(char *s) {char c,*t=s;while(c=gc(),c<=32);*s++=c;while(c=gc(),c>32)*s++=c;return s-t;}
	in void ot() {fwrite(sr,1,C+1,stdout);C=-1;}
	in void flush() {if(C>1<<22) ot();}
	template <typename T>
	in void write(T x,char t)
	{
		int y=0;if(x<0)y=1,x=-x;while(z[++Z]=x%10+48,x/=10);
		if(y)z[++Z]='-';while(sr[++C]=z[Z],--Z);sr[++C]=t;flush();
	}
	in void write(char *s) {int l=strlen(s);for(int i=0;i<l;i++)sr[++C]=*s++;sr[++C]='\n';flush();}
}
using namespace fast_io;
#define pb push_back 
const int N=1e6+5; 
int n,m;ll a[N],b[N];
ll gcd(ll a,ll b) {return !b?a:gcd(b,a%b);}
int main()
{
	n=read(),m=read();for(int i=1;i<=n;i++) a[i]=read();
	sort(a+1,a+n+1);ll st=0;
	for(int i=2;i<=n;i++) st=gcd(a[i]-a[1],st);
	for(int i=1;i<=m;i++) write(gcd(a[1]+read(),st)," \n"[i==m]);
	
	return ot(),0;
}
//Author: disangan233
### Prompt

Construct a CPP code solution to the problem outlined:

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<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define in inline
#define ld long db
namespace fast_io
{
	char buf[1<<12],*p1=buf,*p2=buf,sr[1<<23],z[23],nc;int C=-1,Z=0,Bi=0,ny;
	in char gc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<12,stdin),p1==p2)?EOF:*p1++;}
	in ll read()
	{
		ll x=0;ny=1;while(nc=gc(),(nc<48||nc>57)&&nc!=EOF)if(nc==45)ny=-1;Bi=1;if(nc<0)return nc;
		x=nc-48;while(nc=gc(),47<nc&&nc<58&&nc!=EOF)x=(x<<3)+(x<<1)+(nc^48),Bi++;return x*ny;
	}
	in db gf() {int a=read(),y=ny,b=(nc!='.')?0:read();return (b?a+(db)b/pow(10,Bi)*y:a);}
	in int gs(char *s) {char c,*t=s;while(c=gc(),c<=32);*s++=c;while(c=gc(),c>32)*s++=c;return s-t;}
	in void ot() {fwrite(sr,1,C+1,stdout);C=-1;}
	in void flush() {if(C>1<<22) ot();}
	template <typename T>
	in void write(T x,char t)
	{
		int y=0;if(x<0)y=1,x=-x;while(z[++Z]=x%10+48,x/=10);
		if(y)z[++Z]='-';while(sr[++C]=z[Z],--Z);sr[++C]=t;flush();
	}
	in void write(char *s) {int l=strlen(s);for(int i=0;i<l;i++)sr[++C]=*s++;sr[++C]='\n';flush();}
}
using namespace fast_io;
#define pb push_back 
const int N=1e6+5; 
int n,m;ll a[N],b[N];
ll gcd(ll a,ll b) {return !b?a:gcd(b,a%b);}
int main()
{
	n=read(),m=read();for(int i=1;i<=n;i++) a[i]=read();
	sort(a+1,a+n+1);ll st=0;
	for(int i=2;i<=n;i++) st=gcd(a[i]-a[1],st);
	for(int i=1;i<=m;i++) write(gcd(a[1]+read(),st)," \n"[i==m]);
	
	return ot(),0;
}
//Author: disangan233
```