Problem: There is a grid that consists of W \times H cells. The upper-left-most cell is (1, 1). You are standing on the cell of (1,1) and you are going to move to cell of (W, H). You can only move to adjacent lower-left, lower or lower-right cells.

There are obstructions on several cells. You can not move to it. You cannot move out the grid, either. Write a program that outputs the number of ways to reach (W,H) modulo 1,000,000,009. You can assume that there is no obstruction at (1,1).



Input

The first line contains three integers, the width W, the height H, and the number of obstructions N. (1 \leq W \leq 75, 2 \leq H \leq 10^{18}, 0 \leq N \leq 30) Each of following N lines contains 2 integers, denoting the position of an obstruction (x_i, y_i).

The last test case is followed by a line containing three zeros.

Output

For each test case, print its case number and the number of ways to reach (W,H) modulo 1,000,000,009.

Example

Input

2 4 1
2 1
2 2 1
2 2
0 0 0


Output

Case 1: 4
Case 2: 0

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<string>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> P;
typedef pair<ll,P> P1;

#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define rep(i,x) for(int i=0;i<x;i++)
#define rep1(i,x) for(int i=1;i<=x;i++)
#define rrep(i,x) for(int i=x-1;i>=0;i--)
#define rrep1(i,x) for(int i=x;i>0;i--)
#define sor(v) sort(v.begin(),v.end())
#define rev(s) reverse(s.begin(),s.end())
#define lb(vec,a) lower_bound(vec.begin(),vec.end(),a)
#define ub(vec,a) upper_bound(vec.begin(),vec.end(),a)
#define uniq(vec) vec.erase(unique(vec.begin(),vec.end()),vec.end())
#define mp1(a,b,c) P1(a,P(b,c))

const int INF=1000000000;
const int dir_4[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
const int dir_8[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};

const ll M = 1000000009;

const int matrix_siz = 75;
struct sq_matrix{
	ll n;
	ll a[matrix_siz][matrix_siz];
	void init(int n_){
		n = n_;
		rep(i,n){
			rep(j,n){
				if(i == j)a[i][j] = 1;
				else a[i][j] = 0;
			}
		}
	}
	sq_matrix operator * (const sq_matrix y) {
		sq_matrix ret; ret.n = this->n;
		rep(i,this->n){
			rep(j,this->n){
				ret.a[i][j] = 0;
				rep(k,this->n){
					ret.a[i][j] += this->a[i][k] * y.a[k][j];
					ret.a[i][j] %= M;
				}
			}
		}
		return ret;
	}
};

sq_matrix sq_matrix_powers[62];
void sq_matrix_powers_init(sq_matrix x,int n){
	sq_matrix_powers[0] = x;
	rep1(i,61)sq_matrix_powers[i] = sq_matrix_powers[i-1]*sq_matrix_powers[i-1];
}
sq_matrix sq_matrix_pow(ll k){
	sq_matrix ret; ret.init(sq_matrix_powers[0].n);
	rep(i,61)if((k>>(ll)i)%2 == 1)ret = ret * sq_matrix_powers[i];
	return ret;
}

int main(){
	int test = 0;
	while(1){
		test ++;
		ll w,h,n;
		vector<P> vec;
		scanf("%lld%lld%lld",&w,&h,&n);
		if(w == 0)break;
		else printf("Case %d: ",test);
		rep(i,n){
			ll a,b;
			scanf("%lld%lld",&a,&b);
			if(b != 1)vec.pb( P(b,a-1) );
		}
		sor(vec);
		
		sq_matrix a; a.n = w;
		rep(i,w){
			rep(j,w){
				if(abs(i-j) <= 1)a.a[i][j] = 1;
				else a.a[i][j] = 0;
			}
		}
		sq_matrix_powers_init(a,w);
		sq_matrix ret; ret.init(w);
		ll loc = 1;
		rep(i,vec.size()){
			ret = ret*sq_matrix_pow(vec[i].fr-1-loc);
			loc = vec[i].fr;
			sq_matrix b = a;
			while(i < vec.size() && vec[i].fr == loc){
				rep(j,w)b.a[j][vec[i].sc] = 0;
				i ++;
			}
			i --;
			ret = ret*b;
		}
		ret = ret*sq_matrix_pow(h-loc);
		cout << ret.a[0][w-1] << endl;
	}
}