You are given a tree of n vertices. You are to select k (not necessarily distinct) simple paths in such a way that it is possible to split all edges of the tree into three sets: edges not contained in any path, edges that are a part of exactly one of these paths, and edges that are parts of all selected paths, and the latter set should be non-empty.

Compute the number of ways to select k paths modulo 998244353.

The paths are enumerated, in other words, two ways are considered distinct if there are such i (1 ≤ i ≤ k) and an edge that the i-th path contains the edge in one way and does not contain it in the other.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 10^{5}) — the number of vertices in the tree and the desired number of paths.

The next n - 1 lines describe edges of the tree. Each line contains two integers a and b (1 ≤ a, b ≤ n, a ≠ b) — the endpoints of an edge. It is guaranteed that the given edges form a tree.

Output

Print the number of ways to select k enumerated not necessarily distinct simple paths in such a way that for each edge either it is not contained in any path, or it is contained in exactly one path, or it is contained in all k paths, and the intersection of all paths is non-empty. 

As the answer can be large, print it modulo 998244353.

Examples

Input

3 2
1 2
2 3


Output

7


Input

5 1
4 1
2 3
4 5
2 1


Output

10


Input

29 29
1 2
1 3
1 4
1 5
5 6
5 7
5 8
8 9
8 10
8 11
11 12
11 13
11 14
14 15
14 16
14 17
17 18
17 19
17 20
20 21
20 22
20 23
23 24
23 25
23 26
26 27
26 28
26 29


Output

125580756

Note

In the first example the following ways are valid：

  * ((1,2), (1,2)), 
  * ((1,2), (1,3)), 
  * ((1,3), (1,2)), 
  * ((1,3), (1,3)), 
  * ((1,3), (2,3)), 
  * ((2,3), (1,3)), 
  * ((2,3), (2,3)). 



In the second example k=1, so all n ⋅ (n - 1) / 2 = 5 ⋅ 4 / 2 = 10 paths are valid.

In the third example, the answer is ≥ 998244353, so it was taken modulo 998244353, don't forget it!
#include <bits/stdc++.h>
using namespace std;
const int MAXL = 1048576;
const int P = 998244353;
const int g[] = {3, (P + 1) / 3};
template <typename T1, typename T2>
void Inc(T1 &a, T2 b) {
  a += b;
  if (a >= P) a -= P;
}
template <typename T1, typename T2>
void Dec(T1 &a, T2 b) {
  a -= b;
  if (a < 0) a += P;
}
template <typename T1, typename T2>
T1 Add(T1 a, T2 b) {
  a += b;
  return a >= P ? a - P : a;
}
template <typename T1, typename T2>
T1 Sub(T1 a, T2 b) {
  a -= b;
  return a < 0 ? a + P : a;
}
long long ksm(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, (a *= a) %= P)
    if (b & 1) (ret *= a) %= P;
  return ret;
}
namespace NTT_namespace {
int rev[MAXL];
long long inv[MAXL], G[2][MAXL];
void preworks() {
  inv[1] = 1;
  for (int i = 2; i < MAXL; i++) inv[i] = Sub(P, P / i) * inv[P % i] % P;
  G[0][0] = G[1][0] = 1;
  G[0][1] = ksm(g[0], (P - 1) / MAXL);
  G[1][1] = ksm(g[1], (P - 1) / MAXL);
  for (int i = 2; i < MAXL; i++)
    G[0][i] = G[0][i - 1] * G[0][1] % P, G[1][i] = G[1][i - 1] * G[1][1] % P;
}
int init(int n) {
  int tot = 1, lg2 = 0;
  while (tot < n) tot <<= 1, lg2++;
  for (int i = 0; i < tot; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg2 - 1));
  return tot;
}
void NTT(long long *a, int n, int f) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  long long *ed = a + n;
  for (int i = 2; i <= n; i <<= 1)
    for (long long *x = a, *y = a + (i >> 1); x != ed; x += i, y += i)
      for (int k = 0; k < (i >> 1); k++) {
        long long u = x[k], t = y[k] * G[f][MAXL / i * k] % P;
        x[k] = Add(u, t);
        y[k] = Sub(u, t);
      }
  if (f)
    for (int i = 0; i < n; i++) (a[i] *= inv[n]) %= P;
}
}  // namespace NTT_namespace
struct Poly {
  vector<long long> v;
  Poly() {}
  Poly(int n) { v.resize(n); }
  Poly(const initializer_list<long long> &T) { v = T; }
  long long &operator[](int x) { return v[x]; }
  const long long &operator[](int x) const { return v[x]; }
  int size() const { return v.size(); }
  void resize(int x) { v.resize(x); }
  void reduce() {
    while (v.size() && v.back() == 0) v.pop_back();
  }
  void reverse() { std::reverse(v.begin(), v.end()); }
  void print() const {
    for (size_t i = 0; i < v.size(); i++)
      printf("%lld%c", v[i], "\n "[i != v.size() - 1]);
  }
  friend Poly operator+(const Poly &lhs, const Poly &rhs) {
    if (lhs.size() > rhs.size()) {
      Poly Ret(lhs.size());
      for (int i = 0; i < rhs.size(); i++) Ret[i] = Add(lhs[i], rhs[i]);
      for (int i = rhs.size(); i < lhs.size(); i++) Ret[i] = lhs[i];
      return Ret;
    } else {
      Poly Ret(rhs.size());
      for (int i = 0; i < lhs.size(); i++) Ret[i] = Add(lhs[i], rhs[i]);
      for (int i = lhs.size(); i < rhs.size(); i++) Ret[i] = rhs[i];
      return Ret;
    }
  }
  friend Poly operator-(const Poly &lhs, const Poly &rhs) {
    if (lhs.size() > rhs.size()) {
      Poly Ret(lhs.size());
      for (int i = 0; i < rhs.size(); i++) Ret[i] = Sub(lhs[i], rhs[i]);
      for (int i = rhs.size(); i < lhs.size(); i++) Ret[i] = lhs[i];
      return Ret;
    } else {
      Poly Ret(rhs.size());
      for (int i = 0; i < lhs.size(); i++) Ret[i] = Sub(lhs[i], rhs[i]);
      for (int i = lhs.size(); i < rhs.size(); i++) Ret[i] = Sub(0, rhs[i]);
      return Ret;
    }
  }
  friend Poly operator*(const Poly &lhs, const Poly &rhs) {
    using namespace NTT_namespace;
    static long long t1[MAXL], t2[MAXL];
    if (lhs.size() < 20 || rhs.size() < 20) {
      Poly Ret(lhs.size() + rhs.size() - 1);
      for (int i = 0; i < lhs.size(); i++)
        for (int j = 0; j < rhs.size(); j++) Ret[i + j] += lhs[i] * rhs[j] % P;
      for (long long &v : Ret.v) v %= P;
      return Ret;
    }
    int tot = init(lhs.size() + rhs.size()), siz = lhs.size() + rhs.size() - 1;
    Poly Ret(siz);
    memset(t1, 0, sizeof(long long) * tot);
    memset(t2, 0, sizeof(long long) * tot);
    for (int i = 0; i < lhs.size(); i++) t1[i] = lhs[i];
    for (int i = 0; i < rhs.size(); i++) t2[i] = rhs[i];
    NTT(t1, tot, 0);
    NTT(t2, tot, 0);
    for (int i = 0; i < tot; i++) (t1[i] *= t2[i]) %= P;
    NTT(t1, tot, 1);
    for (int i = 0; i < siz; i++) Ret[i] = t1[i];
    return Ret;
  }
  friend Poly operator*(const Poly &lhs, const long long &rhs) {
    Poly Ret(lhs.size());
    for (int i = 0; i < lhs.size(); i++) Ret[i] = lhs[i] * rhs % P;
    return Ret;
  }
};
const int MAXN = 2E5 + 10;
int n, k, siz[MAXN];
long long fac[MAXN], ifac[MAXN], f[MAXN], sf[MAXN], Ans;
vector<int> T[MAXN];
long long C(int n, int m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return fac[n] * ifac[n - m] % P * ifac[m] % P;
}
pair<Poly, Poly> solve(int l, int r, vector<long long> &a,
                       vector<long long> &b) {
  if (l == r) {
    Poly f(2), g(1);
    f[1] = a[l - 1], f[0] = 1;
    g[0] = b[l - 1];
    return make_pair(f, g);
  }
  int mid = (l + r) >> 1;
  auto Lhs = solve(l, mid, a, b), Rhs = solve(mid + 1, r, a, b);
  return make_pair(Lhs.first * Rhs.first,
                   Lhs.first * Rhs.second + Lhs.second * Rhs.first);
}
void dfs(int x, int fa) {
  long long tmp = 0;
  siz[x] = 1;
  vector<long long> a, b;
  for (int v : T[x])
    if (v != fa) {
      dfs(v, x);
      siz[x] += siz[v];
      Inc(sf[x], sf[v]);
      a.push_back(siz[v]);
      b.push_back(sf[v]);
      Inc(Ans, sf[v] * tmp % P);
      Inc(tmp, sf[v]);
    }
  if (a.size()) {
    auto t = solve(1, a.size(), a, b);
    t.second = t.second * Poly({1, n - siz[x]});
    for (int i = 0; i < t.first.size(); i++)
      Inc(f[x], C(k, k - i) * t.first[i] % P * fac[i] % P);
    for (int i = 0; i < t.second.size(); i++)
      Inc(Ans, C(k, k - i) * t.second[i] % P * fac[i] % P);
  } else
    Inc(f[x], C(k, k));
  Inc(sf[x], f[x]);
}
int main() {
  NTT_namespace::preworks();
  scanf("%d%d", &n, &k);
  if (k == 1)
    return printf("%lld\n", 1ll * n * (n - 1) % P * ksm(2, P - 2) % P), 0;
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= k; i++) fac[i] = fac[i - 1] * i % P;
  ifac[k] = ksm(fac[k], P - 2);
  for (int i = k - 1; i; i--) ifac[i] = ifac[i + 1] * (i + 1) % P;
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), T[u].push_back(v), T[v].push_back(u);
  dfs(1, 0);
  printf("%lld\n", Ans);
}
