Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 2
2 3
1 4

Sample Output 1

3

The i-th robot from the left will be called Robot i, and the j-th exit from the left will be called Exit j.
There are three possible combinations of exits (the exit used by Robot 1, the exit used by Robot 2) as follows:

(Exit 1, Exit 1)

(Exit 1, Exit 2)

(Exit 2, Exit 2)

Sample Input 2

3 4
2 5 10
1 3 7 13

Sample Output 2

8

The exit for each robot can be chosen independently, so there are 2^3 = 8 possible combinations of exits.

Sample Input 3

4 1
1 2 4 5
3

Sample Output 3

1

Every robot uses Exit 1.

Sample Input 4

4 5
2 5 7 11
1 3 6 9 13

Sample Output 4

6

Sample Input 5

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

Sample Output 5

22

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<functional>
#include<cmath>
#include<vector>
//using namespace std;
using std::min;
using std::max;
using std::swap;
using std::sort;
using std::reverse;
using std::random_shuffle;
using std::lower_bound;
using std::upper_bound;
using std::unique;
using std::vector;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef std::pair<int,int> pii;
typedef std::pair<ll,ll> pll;
void open(const char *s){
#ifndef ONLINE_JUDGE
	char str[100];sprintf(str,"%s.in",s);freopen(str,"r",stdin);sprintf(str,"%s.out",s);freopen(str,"w",stdout);
#endif
}
int rd(){int s=0,c,b=0;while(((c=getchar())<'0'||c>'9')&&c!='-');if(c=='-'){c=getchar();b=1;}do{s=s*10+c-'0';}while((c=getchar())>='0'&&c<='9');return b?-s:s;}
void put(int x){if(!x){putchar('0');return;}static int c[20];int t=0;while(x){c[++t]=x%10;x/=10;}while(t)putchar(c[t--]+'0');}
int upmin(int &a,int b){if(b<a){a=b;return 1;}return 0;}
int upmax(int &a,int b){if(b>a){a=b;return 1;}return 0;}
const ll p=1000000007;
const int N=100010;
int cmp(pii a,pii b)
{
	if(a.first!=b.first)
		return a.first<b.first;
	return a.second>b.second;
}
int n,m,t,t2;
int a[N],b[N],d[N];
pii c[N];
ll e[N];
ll add(ll a,ll b)
{
	a+=b;
	return a>=p?a-p:a;
}
void add(int x,ll v)
{
	for(;x<=t2;x+=x&-x)
		e[x]=add(e[x],v);
}
ll sum(int x)
{
	ll s=0;
	for(;x;x-=x&-x)
		s=add(s,e[x]);
	return s;
}
int main()
{
	open("arc101f");
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=m;i++)
		scanf("%d",&b[i]);
	for(int i=1;i<=n;i++)
		if(a[i]>b[1]&&a[i]<b[m])
		{
			int x=lower_bound(b+1,b+m+1,a[i])-b;
			if(b[x]==a[i])
				continue;
			c[++t]=pii(a[i]-b[x-1],b[x]-a[i]);
			d[++t2]=b[x]-a[i];
		}
	sort(d+1,d+t2+1);
	t2=unique(d+1,d+t2+1)-d-1;
	for(int i=1;i<=t;i++)
		c[i].second=lower_bound(d+1,d+t2+1,c[i].second)-d;
	sort(c+1,c+t+1,cmp);
	t=unique(c+1,c+t+1)-c-1;
	ll ans=1;
	for(int i=1;i<=t;i++)
	{
		ll s=sum(c[i].second-1)+1;
		ans+=s;
		add(c[i].second,s);
	}
	ans%=p;
	printf("%lld\n",ans);
	return 0;
}