#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define fi first #define se second #define rep(i,n) for(int i = 0; i < n; ++i) #define rrep(i,n) for(int i = 1; i <= n; ++i) #define drep(i,n) for(int i = n-1; i >= 0; --i) #define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next) #define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++) #define rng(a) a.begin(),a.end() #define maxs(x,y) x = max(x,y) #define mins(x,y) x = min(x,y) #define pb push_back #define sz(x) (int)(x).size() #define pcnt __builtin_popcount #define snuke srand((unsigned)clock()+(unsigned)time(NULL)); using namespace std; typedef long long int ll; typedef pair P; inline int in() { int x; scanf("%d",&x); return x;} const int MX = 100005, INF = 1000010000; const ll LINF = 1000000000000000000ll; const double eps = 1e-10; const int di[] = {-1,0,1,0}, dj[] = {0,-1,0,1}; //^ // Mod int const int mod = 1000000007; struct mint{ ll x; mint():x(0){} mint(ll x):x((x%mod+mod)%mod){} mint operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;} mint operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;} mint operator*=(const mint& a){ (x*=a.x)%=mod; return *this;} mint operator+(const mint& a)const{ return mint(*this) += a;} mint operator-(const mint& a)const{ return mint(*this) -= a;} mint operator*(const mint& a)const{ return mint(*this) *= a;} bool operator==(const mint& a)const{ return x == a.x;} }; // typedef vector vi; typedef vector vm; typedef vector vvm; int n, k; vi to[MX]; int s[MX]; vvm dp[MX]; void dfs(int v, int p=-1) { s[v] = 1; dp[v] = vvm(2,vm(2)); dp[v][0][0] = 1; dp[v][1][1] = 1; for (int u : to[v]) { if (u == p) continue; dfs(u,v); vvm a = dp[v]; vvm& b = dp[u]; rep(i,2) dp[v][i] = vm(s[v]+s[u]+1,0); rep(i,s[v]+1)rep(j,s[u]+1) { dp[v][0][i+j] += a[0][i]*b[0][j]; dp[v][0][i+j] += a[0][i]*b[1][j]; dp[v][1][i+j] += a[1][i]*b[1][j]; } s[v] += s[u]; } } int main(){ cin >> n >> k; rep(i,n-1) { int a, b; cin >> a >> b; to[a].pb(b); to[b].pb(a); } dfs(0); mint ans = 0; rep(i,2) ans += dp[0][i][k]; cout << ans.x << endl; return 0; }