CodeForces 559C Gerald and Giant Chess

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

3 4 2
2 2
2 3

Output

2

Input

100 100 3
15 16
16 15
99 88

Output

545732279

总路径数减去经过黑点的路径数就是答案。若之前无障碍，走到点（x,y）的路径数=C(x,x+y)。实际计算时为避免重复，要减去从之前的黑点到当前黑点的路径数。

计算的值非常大，由于涉及取模除法，需要用到乘法逆元算法。

乘法逆元什么鬼！虽然看懂了但是不会实现，我选择死亡

高端解析和高端代码链接：http://blog.csdn.net/sdfzyhx/article/details/51822192

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int mxn=200200;
const long long p=1000000007;
long long jc[mxn],rc[mxn];//阶乘 逆元 
long long dp[2005];
int h,w,n;
struct node{
    int x,y;
}a[2005];
int cmp(node a,node b){
    if(a.x!=b.x)return a.x<b.x;
    return a.y<b.y;
}
void eu(long long a,long long b,long long &x,long long &y){//扩展欧几里得 
    if(!b){
        x=1;y=0;
    }
    else{
        eu(b,a%b,y,x);
        y-=x*(a/b);
    }
    return;
}
void init(){
    int i;
    jc[0]=rc[0]=1;
    long long x;
    for(i=1;i<200005;i++){
        jc[i]=(jc[i-1]*i)%p;//阶乘
        eu(jc[i],p,rc[i],x);
        rc[i]=(rc[i]%p+p)%p;
    }
    return;
}
long long c(int x,int y){
    return jc[y]*rc[x]%p*rc[y-x]%p;
}
int main(){
    init();
    scanf("%d%d%d",&h,&w,&n);
    int i,j;
    for(i=1;i<=n;i++){
        scanf("%d%d",&a[i].x,&a[i].y);
    }
    n++;a[n].x=h;a[n].y=w;
    sort(a+1,a+n+1,cmp);
    for(i=1;i<=n;i++){
        dp[i]=c(a[i].x-1,a[i].x+a[i].y-2);
//        printf("---%d---
",dp[i]);
        for(j=1;j<i;j++){
            if(a[j].y<=a[i].y)
                dp[i]=((dp[i]-dp[j]*c(a[i].x-a[j].x,a[i].x+a[i].y-a[j].x-a[j].y)%p)+p)%p;//减去重复值 
        }
    }
    printf("%lld
",dp[n]);
    return 0;
}