Problem:
Three-way Branch

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.

Sample Input

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

Output for the Sample Input

Case 1: 4
Case 2: 0

Solution in C++ (judge verdict: Accepted)

#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000009
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(kbrni,n)cout<<" "<<a[kbrni];cout<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define svecp(v) cout<<#v<<":";each(kbrni,v)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

template<typename T> class mat : public vector<vector<T> > {
private:
    int r,c;    //行,列
public:
    int row() const {
        return r;
    }
    int column() const {
        return c;
    }
    mat(int n,int m,T val = 0){
        this->r = n,this->c = m;
        rep(i,n){
            this->push_back(vector<T>(m,val));
        }
    }
    mat operator+(const mat& another){
        if(this->r != another.r && this->c != another.c){
            cout << "足し算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] + another[i][j];
            }
        }
        return X;
    }
    mat operator+(const T val){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] + val;
            }
        }
        return X;
    }
    mat operator-(const mat& another){
        if(this->r != another.r && this->c != another.c){
            cout << "引き算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] - another[i][j];
            }
        }
        return X;
    }
    mat operator-(const T val){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] - val;
            }
        }
        return X;
    }
    vector<T> operator*(const vector<T>& another){
        if(this->c != (int)another.size()){
            cout << "掛け算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        vector<T> vec(this->r,0);
        rep(i,this->r){
            rep(j,this->c){
                vec[i] += (*this)[i][j] * another[j];
            }
        }
        return vec;
    }
    mat operator*(const mat& another){
        if(this->c != another.r){
            cout << "掛け算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,another.c);
        rep(i,this->r){
            rep(k,this->c){
                rep(j,another.c){
                    X[i][j] = (X[i][j] + (ll)(*this)[i][k]*another[k][j]%MOD)%MOD;
                }
            }
        }
        return X;
    }
    mat operator-(){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = -(*this)[i][j];
            }
        }
        return X;
    }
    void print(){
        rep(i,this->r){
            rep(j,(this->c)-1){
                cout << (*this)[i][j] << ",";
            }
            cout << (*this)[i][(this->c)-1] << endl;
        }
    }
};

template<typename T> mat<T> pow(mat<T> A,ll cnt)
{
    if(A.row() != A.column()){
        cout << "累乗不可" << endl;
    }
    int n = A.row();
	mat<T> B(n,n);
	rep(i,n){
		B[i][i] = 1;
	}
	while(cnt>0){
		if(cnt & 1){
			B = B*A;
		}
		A = A*A;
		cnt >>= 1;
	}
	return B;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int cnt = 1;
    while(1){
        int w,n;
        ll h;
        cin >> w >> h >> n;
        if(w == 0){
            break;
        }
        mat<int> A(w,w);
        vector<pll> vec;
        rep(i,n){
            ll a,b;
            cin >> a >> b;
            if(b == 1){
                continue;
            }
            vec.pb(pll(b-1,a-1));
        }
        n = len(vec);
        sort(all(vec));
        rep(i,w){
            srep(j,max(i-1,0),min(i+2,w)){
                A[i][j] = 1;
            }
        }
        int id = 0;
        ll pos = 0;
        mat<int> Ans(w,w);
        rep(i,w){
            Ans[i][i] = 1;
        }
        while(id < n){
            mat<int> B = pow(A,vec[id].fi-1-pos);
            Ans = Ans*B;
            B = A;
            pos = vec[id].fi;
            while(id < n){
                rep(i,w){
                    B[i][vec[id].se] = 0;
                }
                id++;
                if(id < n && vec[id-1].fi != vec[id].fi){
                    break;
                }
            }
            Ans = Ans*B;
        }
        mat<int> B = pow(A,h-1-pos);
        Ans = Ans*B;
        cout << "Case " << cnt << ": " << Ans[0][w-1] << "\n";
        cnt++;
    }
    return 0;
}