CF559C Gerald and Giant Chess

题意

C. Gerald and Giant Chess

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Giant chess is quite common in Geraldion. We will not delve into the rules of the game, we'll just say that the game takes place on an h × w field, and it is painted in two colors, but not like in chess. Almost all cells of the field are white and only some of them are black. Currently Gerald is finishing a game of giant chess against his friend Pollard. Gerald has almost won, and the only thing he needs to win is to bring the pawn from the upper left corner of the board, where it is now standing, to the lower right corner. Gerald is so confident of victory that he became interested, in how many ways can he win?

The pawn, which Gerald has got left can go in two ways: one cell down or one cell to the right. In addition, it can not go to the black cells, otherwise the Gerald still loses. There are no other pawns or pieces left on the field, so that, according to the rules of giant chess Gerald moves his pawn until the game is over, and Pollard is just watching this process.

Input

The first line of the input contains three integers: h, w, n — the sides of the board and the number of black cells (1 ≤ h, w ≤ 10⁵, 1 ≤ n ≤ 2000).

Next n lines contain the description of black cells. The i-th of these lines contains numbers r_i, c_i (1 ≤ r_i ≤ h, 1 ≤ c_i ≤ w) — the number of the row and column of the i-th cell.

It is guaranteed that the upper left and lower right cell are white and all cells in the description are distinct.

Output

Print a single line — the remainder of the number of ways to move Gerald's pawn from the upper left to the lower right corner modulo 10⁹ + 7.

Examples

Input

Copy

3 4 2
2 2
2 3

Output

Copy

2

Input

Copy

100 100 3
15 16
16 15
99 88

Output

Copy

545732279

给出一个h*w的棋盘(h,w<=1e5),其中有n个位置不能走(n<=2000),现在要从左上角走到右下角，每步只能向下或者向右走一步。问有多少种走法？右下角保证可以走到。

分析

参照PoemK的题解。

对于右走x步，下走y步的无限制方案数是C(x+y,y)，可以记为C(x,y)。

首先将最右下角也作为一个不能走的位置,最后要求出到达这个位置的合法路径数dp[n]。

对所有不能走的位置排序，先比较行，再比较列。

对于位置i，首先有dp[i]=C(x[i],y[i]);

然后到达i途中不能经过任何一个障碍位置，所以枚举第一个障碍位置。dp[i]=dp[i]−∑dp[k]∗C(k−>i)(ifC(k−>i)>0)

时间复杂度\(O(n^2)\)

代码

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=2e5,mod=1e9+7;
int add(int x,int y) {return (x+=y)>=mod?x-mod:x;}
int mul(int x,int y) {return (ll)x*y%mod;}
int fpow(int x,int k){
	int re=1;
	for(;k;k>>=1,x=mul(x,x))
		if(k&1) re=mul(re,x);
	return re;
}
int fac[N],ifac[N];
int binom(int n,int m) {return mul(fac[n],mul(ifac[m],ifac[n-m]));}
#define x first
#define y second
pair<int,int> a[N];
int h,w,n,f[N];
int main(){
	fac[0]=ifac[0]=1;
	for(int i=1;i<N;++i){
		fac[i]=mul(fac[i-1],i);
		ifac[i]=fpow(fac[i],mod-2);
	}
	read(h),read(w),read(n);
	for(int i=1;i<=n;++i) read(a[i].x),read(a[i].y);
	sort(a+1,a+n+1);
	a[n+1].x=h,a[n+1].y=w;
	for(int i=1;i<=n+1;++i){
		f[i]=binom(a[i].x+a[i].y-2,a[i].x-1);
		for(int j=1;j<i;++j)if(a[j].x<=a[i].x&&a[j].y<=a[i].y)
			f[i]=add(f[i],mod-mul(f[j],binom(a[i].x+a[i].y-a[j].x-a[j].y,a[i].x-a[j].x)));
	}
	printf("%d\n",f[n+1]);
	return 0;
}