/*
Author: AquaBlaze
Time: 2021-01-03 13:25:24
Generated by powerful Codeforces Tool :^)
You can download the binary file in here https://github.com/xalanq/cf-tool (Windows, macOS, Linux)
Keqing best girl :)
Nephren will always be in my heart
*/
#include<bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define eb emplace_back
#define all(a) (a).begin(),(a).end()
#define SZ(a) (int)(a).size()
#define FOR(i, a, b) for(int i=(a); i<=(b); ++i)
#define ROF(i, a, b) for(int i=(a); i>=(b); --i)
#define make_unique(a) sort(all((a))), (a).resize(unique(all((a)))-(a).begin())
#define pc(x) putchar(x)
#define MP make_pair
#define MT make_tuple
using namespace std;
typedef long long i64;
typedef tuple<int,int,int> iii;
typedef pair<int,int> pii;
typedef pair<i64,i64> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
const int N = 1005;
const int mod = 1000000007;
int add(int a, int b){ return ((a+=b)>=mod)?a-mod:a; }
void adding(int &a, int b){ a = add(a, b); }
int mul(int a, int b){ return a*1ll*b%mod; }
pii operator + (const pii& A, const pii& B){
return pii(add(A.x,B.x), add(A.y,B.y));
}
pii operator - (const pii& A, const pii& B){
return pii(add(A.x,mod-B.x), add(A.y,mod-B.y));
}
vector<int> g[N];
int result[N];
pii dp[N][2];
void add_node(int u, int v){
result[u] -= !result[v];
if(result[u]){
dp[u][1] = dp[u][1]+dp[v][0]+dp[v][1];
}else{
dp[u][0] = dp[u][0]+dp[v][1];
dp[u][1] = dp[u][1]+dp[v][0];
}
result[u] += !result[v];
}
void adjust(int u){
if(!result[u]){
dp[u][0] = dp[u][0]+pii(1, 0);
dp[u][1] = dp[u][1]+pii(0, 1);
}else dp[u][1] = dp[u][1]+pii(1, 1);
}
void dfs(int u, int p){
for(const int &e : g[u]){
if(e == p) continue;
dfs(e, u);
result[u] += (!result[e]);
}
adjust(u);
for(const int &e : g[u]){
if(e == p) continue;
add_node(u, e);
}
}
void solve(){
int n;
i64 d;
cin >> n >> d;
FOR(i, 2, n){
int a, b;
cin >> a >> b;
g[a].eb(b);
g[b].eb(a);
}
vvi m(2, vi(2,0));
vi res(2, 0);
ROF(i, n, 1){
memset(dp, 0, sizeof dp);
memset(result, 0, sizeof result);
dfs(i, -1);
int r = !(!result[i]);
res[1-r]++;
FOR(j, 0, 1){
adding(m[j^1][0], dp[i][j].x);
adding(m[j^1][1], dp[i][j].y);
}
}
FOR(t, 2, d){
vi tmp(2, 0);
FOR(i, 0, 1){
FOR(j, 0, 1){
adding(tmp[i], mul(res[j], m[i][j]));
}
}
res = tmp;
}
int ans = 0;
adding(ans, mul(dp[1][1].x, res[0]));
adding(ans, mul(dp[1][1].y, res[1]));
printf("%d",ans);
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
solve();
return 0;
}
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*/
