Problem: There are N robots and M exits on a number line. The N + M coordinates of these are all integers and all distinct. For each i (1 \leq i \leq N), the coordinate of the i-th robot from the left is x_i. Also, for each j (1 \leq j \leq M), the coordinate of the j-th exit from the left is y_j.

Snuke can repeatedly perform the following two kinds of operations in any order to move all the robots simultaneously:

* Increment the coordinates of all the robots on the number line by 1.
* Decrement the coordinates of all the robots on the number line by 1.



Each robot will disappear from the number line when its position coincides with that of an exit, going through that exit. Snuke will continue performing operations until all the robots disappear.

When all the robots disappear, how many combinations of exits can be used by the robots? Find the count modulo 10^9 + 7. Here, two combinations of exits are considered different when there is a robot that used different exits in those two combinations.

Constraints

* 1 \leq N, M \leq 10^5
* 1 \leq x_1 < x_2 < ... < x_N \leq 10^9
* 1 \leq y_1 < y_2 < ... < y_M \leq 10^9
* All given coordinates are integers.
* All given coordinates are distinct.

Input

Input is given from Standard Input in the following format:


N M
x_1 x_2 ... x_N
y_1 y_2 ... y_M


Output

Print the number of the combinations of exits that can be used by the robots when all the robots disappear, modulo 10^9 + 7.

Examples

Input

2 2
2 3
1 4


Output

3


Input

3 4
2 5 10
1 3 7 13


Output

8


Input

4 1
1 2 4 5
3


Output

1


Input

4 5
2 5 7 11
1 3 6 9 13


Output

6


Input

10 10
4 13 15 18 19 20 21 22 25 27
1 5 11 12 14 16 23 26 29 30


Output

22

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#include<stdio.h>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
ll tree[100001],n,m,a[100001],b[100001],t1,t2,d[100001];
pair<ll,ll>c[100001];
ll quyu(ll aa,ll bb)
{
	return (aa+bb)%mod;//防止aa+bb超过mod 
}
void add(ll under,ll v)
{
	for(;under<=n;under+=under&-under)tree[under]=quyu(tree[under],v);//用quyu函数，防超mod 
}
ll sum(ll x)
{
	ll ans=0;
	for(;x;x-=x&-x)ans=quyu(ans,tree[x]);//同上 
	return ans;
}
bool cmp(pair<ll,ll>one,pair<ll,ll>two)//先比较到左边EXIT的距离，再比到右边EXIT的距离。
{
	if(one.first!=two.first)return one.first<two.first; 
	return one.second>two.second;
}
int main()
{
	scanf("%lld %lld",&n,&m);
	for(ll u=1;u<=n;u++)
		scanf("%lld",&a[u]);
	for(ll u=1;u<=m;u++)
		scanf("%lld",&b[u]);
	for(ll u=1;u<=n;u++)
		if(a[u]<b[m]&&a[u]>b[1])//除去在第一个EXIT左边的和最后一个EXIT右边的 
		{
			int xia=lower_bound(b+1,b+m+1,a[u])-b;
			if(a[u]==b[xia])continue;//在EXIT中的不管。 
			c[++t1]=make_pair(a[u]-b[xia-1],b[xia]-a[u]);//存到左边EXIT和到右边EXIT的距离和 
			d[++t2]=b[xia]-a[u];//单纯记录到右边EXIT的距离 
		}
	sort(d+1,d+t2+1);
	t2=unique(d+1,d+t2+1)-d-1;//排序再去重 
	for(ll u=1;u<=t1;u++)
		c[u].second=lower_bound(d+1,d+t2+1,c[u].second)-d;
	sort(c+1,c+t1+1,cmp);
	t1=unique(c+1,c+t1+1)-c-1;//同上 
	ll ans=1;
	for(ll u=1;u<=t1;u++)//维护树状数组 
	{
		ll s=sum(c[u].second-1)+1;//记得+1 
		ans+=s;//累加 
		add(c[u].second,s);//插入 
	}
	ans%=mod;
	printf("%lld",ans);
	return 0;
}