// SNAKE AND LADDER // Official Solution by Jacob Plachta #define DEBUG 0 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; i=0; i--) #define FoxR1(i,n) for (i=n; i>0; i--) #define FoxRI(i,a,b) for (i=b; i>=a; i--) #define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++) #define Min(a,b) a=min(a,b) #define Max(a,b) a=max(a,b) #define Sz(s) int((s).size()) #define All(s) (s).begin(),(s).end() #define Fill(s,v) memset(s,v,sizeof(s)) #define pb push_back #define mp make_pair #define x first #define y second template T Abs(T x) { return(x<0 ? -x : x); } template T Sqr(T x) { return(x*x); } string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } const int INF = (int)1e9; const LD EPS = 1e-9; const LD PI = acos(-1.0); //#if DEBUG #define GETCHAR getchar /*#else #define GETCHAR getchar_unlocked #endif*/ bool Read(int &x) { char c,r=0,n=0; x=0; for(;;) { c=GETCHAR(); if ((c<0) && (!r)) return(0); if ((c=='-') && (!r)) n=1; else if ((c>='0') && (c<='9')) x=x*10+c-'0',r=1; else if (r) break; } if (n) x=-x; return(1); } bool ReadLL(LL &x) { char c,r=0,n=0; x=0; for(;;) { c=GETCHAR(); if ((c<0) && (!r)) return(0); if ((c=='-') && (!r)) n=1; else if ((c>='0') && (c<='9')) x=x*10+c-'0',r=1; else if (r) break; } if (n) x=-x; return(1); } #define MOD 1000000007 int main() { if (DEBUG) freopen("in.txt","r",stdin); int T,t; LL N; int K; int i,ans; LL a,b,x,y; static pair P[1000]; Read(T); Fox1(t,T) { ReadLL(N),Read(K); Fox(i,K) ReadLL(P[i].x),Read(P[i].y); if (K==2*N-1) { ans=1; goto Done; } sort(P,P+K); a=1,b=N; Fox(i,K-1) if ((P[i].x==P[i+1].x) && (P[i].x==a)) a++; FoxR(i,K-1) if ((P[i].x==P[i+1].x) && (P[i].x==b)) b--; ans=0; Fox(i,K-1) if ((P[i].x==P[i+1].x) && (P[i].x>=a) && (P[i].x<=b)) goto Done; x=-1; Fox(i,K) if ((P[i].x>=a) && (P[i].x<=b)) if (x<0) x=y=P[i].x; else { if ((P[i].x-y)%2==(P[i].y+P[i-1].y)%2) goto Done; y=P[i].x; } if (x<0) { N=b-a+1; if (N==1) ans=2; else { N=(N-1)%MOD; ans=(4+2*N*(N+1))%MOD; } } else { a=max(1LL,x-a); b=max(1LL,b-y); ans=2*(a%MOD)*(b%MOD)%MOD; } Done:; printf("Case #%d: %d\n",t,ans); } return(0); }