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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
const int maxn = 1<<18, mod = 998244353;
using ll = long long;
int n, m;
struct node
{
	ll a, b, sum;
	node() {sum = 0; a = 1;b = 0;}
	node(int val)
	{
		sum = val;
		a = 1;b = 0;
	}
	void apply(ll c, ll d) {
		ll na = c*a%mod;
		ll nb = (c*b + d)%mod;
		a = na, b = nb;
	}
};
 
node temp, broken;
 
node merge(node l, node r)
{
	temp.sum = (l.sum + r.sum)%mod;
	temp.a = 1;temp.b = 0;
	return temp;
}
 
struct segment_tree
{
	node tr[4 * maxn];
 
	void push(int l, int r, int idx)
	{
		tr[idx].sum = ( ((r - l + 1) * tr[idx].b) + (tr[idx].sum * tr[idx].a) )%mod;
		if(l != r)
		{
			tr[2 * idx + 1].apply(tr[idx].a, tr[idx].b);
			tr[2 * idx + 2].apply(tr[idx].a, tr[idx].b);
		}
		tr[idx].a = 1;tr[idx].b = 0;
	}
 
	void init(int l, int r, int idx)
	{
		if(l == r)
		{
			tr[idx] = node();
			return;
		}
 
		int mid = (l + r) >> 1;
		init(l, mid, 2 * idx + 1);
		init(mid + 1, r, 2 * idx + 2);
 
		tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
	}
 
	void update(int qL, int qR, int A, int B, int l, int r, int idx)
	{
		push(l, r, idx);
 
		if(qL > r || l > qR)
			return;
 
		if(qL <= l && r <= qR)
		{
			tr[idx].apply(A, B);
			push(l, r, idx);
			return;
		}
 
		int mid = (l + r) >> 1;
		update(qL, qR, A, B, l, mid, 2 * idx + 1);
		update(qL, qR, A, B, mid + 1, r, 2 * idx + 2);
 
		tr[idx] = merge(tr[2 * idx + 1], tr[2 * idx + 2]);
	}
 
	node query(int qL, int qR, int l, int r, int idx)
	{
		push(l, r, idx);
 
		if(l > qR || r < qL)
			return broken;
 
		if(qL <= l && r <= qR)
			return tr[idx];
 
		int mid = (l + r) >> 1;
		return merge(query(qL, qR, l, mid, 2 * idx + 1), query(qL, qR, mid + 1, r, 2 * idx + 2));
	}
};
 
segment_tree st;
vector<int> a, b;
int tree[maxn];
signed main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	a.resize(n);
	b.resize(m);
	for(auto &i : a) cin >> i;
	for(auto &i : b) cin >> i;
	vector<array<int, 2>> segs;
	vector<int> C{0};
	for(auto &i : a) {
		auto it = upper_bound(all(b), i);
		if(it == b.begin() || it == b.end()) continue;
		auto it2 = it; it2--;
		segs.push_back({i-*it2, *it-i});
		C.push_back(segs.back()[1]);
	}
	sort(all(C));
	for(auto &[x, y] : segs) y = lower_bound(all(C), y)-C.begin();
	for(auto &[x, y] : segs) y *= -1;
	sort(all(segs));
	segs.erase(unique(all(segs)), segs.end());
	for(auto &[x, y] : segs) y *= -1;
	st.init(0, n, 0);
	st.update(0, 0, 0, 1, 0, n, 0);
	for(auto &[cur, p] : segs) {
		st.update(p, p, 1, st.query(0, p-1, 0, n, 0).sum, 0, n, 0);
	}
	cout << st.query(0, n, 0, n, 0).sum << '\n';
}