#define _USE_MATH_DEFINES #include using namespace std; //template #define rep(i,a,b) for(int i=(a);i<(b);i++) #define rrep(i,a,b) for(int i=(a);i>(b);i--) #define ALL(v) (v).begin(),(v).end() typedef long long int ll; typedef pair P; template inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } templatevoid Fill(A(&array)[N],const T &val){fill((T*)array, (T*)(array+N), val);} const int inf = 0x3fffffff; const ll INF = 0x3fffffffffffffff; //template end ll mod=1e9+7; struct Mint { ll val; Mint inv() const{ ll tmp,a=val,b=mod,x=1,y=0; while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y); return Mint(x); } public: Mint():val(0){} Mint(ll x):val(x>=0?x%mod:x%mod+mod){} int mtoi(){return this->val; } Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;} Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;} Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;} Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;} Mint& operator/=(const Mint& x){return *this *= x.inv();} bool operator==(const Mint& x) const{return val == x.val;} bool operator!=(const Mint& x) const{return val != x.val;} bool operator<(const Mint& x) const{return val < x.val;} bool operator<=(const Mint& x) const{return val <= x.val;} bool operator>(const Mint& x) const{return val > x.val;} bool operator>=(const Mint& x) const{return val >= x.val;} Mint operator+(const Mint& x) const{return Mint(*this)+=x;} Mint operator-(const Mint& x) const{return Mint(*this)-=x;} Mint operator*(const Mint& x) const{return Mint(*this)*=x;} Mint operator/(const Mint& x) const{return Mint(*this)/=x;} }; struct factorial { vector Fact, Finv; public: factorial(int maxx){ Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*Mint(i+1); Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*Mint(i); } Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];} Mint nPr(int n,int r){if(n<0||n sum(d+1,0); rep(i,0,d)sum[i+1]=sum[i]+(1<=0&&(l+r-k)%2==0)lca=(l+r-k)/2; if(lca==-1||lca>l||lca>r){ printf("0\n"); return 0; } rep(i,0,d){ int dep=1<