Problem: The Main Martian Tree grows on Mars. It is a binary tree (a rooted tree, with no more than two sons at each vertex) with n vertices, where the root vertex has the number 1. Its fruits are the Main Martian Fruits. It's summer now, so this tree does not have any fruit yet.

Autumn is coming soon, and leaves and branches will begin to fall off the tree. It is clear, that if a vertex falls off the tree, then its entire subtree will fall off too. In addition, the root will remain on the tree. Formally: the tree will have some connected subset of vertices containing the root.

After that, the fruits will grow on the tree (only at those vertices which remain). Exactly x fruits will grow in the root. The number of fruits in each remaining vertex will be not less than the sum of the numbers of fruits in the remaining sons of this vertex. It is allowed, that some vertices will not have any fruits.

Natasha wondered how many tree configurations can be after the described changes. Since this number can be very large, output it modulo 998244353.

Two configurations of the resulting tree are considered different if one of these two conditions is true:

  * they have different subsets of remaining vertices;
  * they have the same subset of remaining vertices, but there is a vertex in this subset where they have a different amount of fruits.

Input

The first line contains two integers: n and x (1 ≤ n ≤ 10^5, 0 ≤ x ≤ 10^{18}) — the size of the tree and the number of fruits in the root.

The i-th of the following (n-1) lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — vertices connected by the i-th edge of the tree.

It is guaranteed that the input data describes a correct binary tree with the root at the vertex 1.

Output

Print one number — the number of configurations of the resulting tree modulo 998244353.

Examples

Input

3 2
1 2
1 3


Output

13


Input

2 5
1 2


Output

7


Input

4 10
1 2
1 3
3 4


Output

441

Note

Consider the first example. <image>

There are 2 fruits at the vertex 1. The following 13 options are possible:

  * there is no vertex 2, there is no vertex 3; <image>
  * there is no vertex 2, there are no fruits at the vertex 3; <image>
  * there is no vertex 2, there is 1 fruit at the vertex 3; <image>
  * there is no vertex 2, there are 2 fruits at the vertex 3; <image>
  * there are no fruits at the vertex 2, there is no vertex 3; <image>
  * there are no fruits at the vertex 2, there are no fruits at the vertex 3; <image>
  * there are no fruits at the vertex 2, there is 1 fruit at the vertex 3; <image>
  * there are no fruits at the vertex 2, there are 2 fruits at the vertex 3; <image>
  * there is 1 fruit at the vertex 2, there is no vertex 3; <image>
  * there is 1 fruit at the vertex 2, there are no fruits at the vertex 3; <image>
  * there is 1 fruit at the vertex 2, there is 1 fruit at the vertex 3; <image>
  * there are 2 fruits at the vertex 2, there is no vertex 3; <image>
  * there are 2 fruits at the vertex 2, there are no fruits at the vertex 3. <image>



Consider the second example. There are 5 fruits at the vertex 1. The following 7 options are possible:

  * there is no vertex 2;
  * there are no fruits at the vertex 2;
  * there is 1 fruit at the vertex 2;
  * there are 2 fruits at the vertex 2;
  * there are 3 fruits at the vertex 2;
  * there are 4 fruits at the vertex 2;
  * there are 5 fruits at the vertex 2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int mod = 998244353;
const int root = 31;
const int LOG = 23;
const int N = 1e5 + 5;
vector<int> G[LOG + 1];
vector<int> rev[LOG + 1];
inline void _add(int &x, int y) {
  if ((x += y) >= mod) {
    x -= mod;
  }
}
inline int _sum(int a, int b) { return a + b < mod ? a + b : a + b - mod; }
inline int _sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline int _mul(int a, int b) { return (1ll * a * b) % mod; }
inline int _binpow(int x, int p) {
  int res = 1;
  int tmp = x;
  while (p > 0) {
    if (p & 1) {
      res = _mul(res, tmp);
    }
    tmp = _mul(tmp, tmp);
    p >>= 1;
  }
  return res;
}
inline int _rev(int x) { return _binpow(x, mod - 2); }
void precalc() {
  for (int start = root, lvl = LOG; lvl >= 0;
       lvl--, start = _mul(start, start)) {
    int tot = 1 << lvl;
    G[lvl].resize(tot);
    for (int cur = 1, i = 0; i < tot; i++, cur = _mul(cur, start)) {
      G[lvl][i] = cur;
    }
  }
  for (int lvl = 1; lvl <= LOG; lvl++) {
    int tot = 1 << lvl;
    rev[lvl].resize(tot);
    for (int i = 1; i < tot; i++) {
      rev[lvl][i] = ((i & 1) << (lvl - 1)) | (rev[lvl][i >> 1] >> 1);
    }
  }
}
void fft(vector<int> &a, int sz, bool invert) {
  int n = 1 << sz;
  for (int j, i = 0; i < n; i++) {
    if ((j = rev[sz][i]) < i) {
      swap(a[i], a[j]);
    }
  }
  for (int f1, f2, lvl = 0, len = 1; len < n; len <<= 1, lvl++) {
    for (int i = 0; i < n; i += (len << 1)) {
      for (int j = 0; j < len; j++) {
        f1 = a[i + j];
        f2 = _mul(a[i + j + len], G[lvl + 1][j]);
        a[i + j] = _sum(f1, f2);
        a[i + j + len] = _sub(f1, f2);
      }
    }
  }
  if (invert) {
    reverse(a.begin() + 1, a.end());
    int rn = _rev(n);
    for (int i = 0; i < n; i++) {
      a[i] = _mul(a[i], rn);
    }
  }
}
vector<int> multiply(const vector<int> &a, const vector<int> &b) {
  vector<int> fa(begin(a), end(a));
  vector<int> fb(begin(b), end(b));
  int n = (int)a.size();
  int m = (int)b.size();
  int maxnm = max(n, m), sz = 0;
  while ((1 << sz) < maxnm) sz++;
  sz++;
  fa.resize(1 << sz);
  fb.resize(1 << sz);
  fft(fa, sz, false);
  fft(fb, sz, false);
  int SZ = 1 << sz;
  for (int i = 0; i < SZ; i++) {
    fa[i] = _mul(fa[i], fb[i]);
  }
  fft(fa, sz, true);
  while ((int)fa.size() > 1 && !fa.back()) fa.pop_back();
  return fa;
}
struct Polynom {
  vector<int> a;
  Polynom() { a.resize(1); }
  Polynom(const vector<int> _a) : a(_a) {}
  int get_size() const { return (int)a.size(); }
  Polynom operator*(const Polynom &o) const {
    return Polynom(multiply(a, o.a));
  }
  Polynom operator+(const Polynom &o) const {
    Polynom res;
    int n = this->get_size();
    int m = o.get_size();
    res.a.resize(max(n, m));
    for (int i = 0; i < n; i++) {
      _add(res.a[i], this->a[i]);
    }
    for (int i = 0; i < m; i++) {
      _add(res.a[i], o.a[i]);
    }
    while ((int)res.a.size() > 1 && !res.a.back()) res.a.pop_back();
    return res;
  }
};
struct Func {
  vector<Polynom> a;
  int len = 0;
  int get_len() const { return len; }
  void add_polynom(const Polynom &add_poly) {
    a.push_back(add_poly);
    len += add_poly.get_size();
    assert(add_poly.get_size());
  }
  pair<Polynom, Polynom> gen_polynom(int l, int r) {
    if (l == r) {
      assert(a[l].get_size());
      return make_pair(a[l], a[l]);
    }
    int m = (l + r) >> 1;
    pair<Polynom, Polynom> L = gen_polynom(l, m);
    pair<Polynom, Polynom> R = gen_polynom(m + 1, r);
    return make_pair((L.first * R.second) + R.first, L.second * R.second);
  }
  Polynom gen_polynom() {
    assert(!a.empty());
    Polynom p = gen_polynom(0, (int)a.size() - 1).first;
    if ((++p.a[0]) == mod) p.a[0] = 0;
    p.a.insert(p.a.begin(), 0);
    return p;
  }
  bool empty() const { return a.empty(); }
};
int n;
long long X;
vector<int> g[N];
int C[N];
Func f[N];
void dfs(int v, int par = -1) {
  int child = 0;
  for (int to : g[v]) {
    if (to == par) continue;
    dfs(to, v);
    child++;
    if (f[to].get_len() > f[v].get_len()) {
      swap(f[v], f[to]);
    }
    if (f[to].empty()) {
      continue;
    }
    f[v].add_polynom(f[to].gen_polynom());
  }
  if (child < 2) {
    f[v].add_polynom(Polynom({0, 1}));
  }
}
int main() {
  precalc();
  scanf("%d %lld", &n, &X);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  {
    fill(C, C + N, 1);
    X %= mod;
    for (int cur = 1, i = 1; i < N; i++) {
      C[i] = _mul(C[i], cur);
      cur = _mul(cur, _sum(X, i));
    }
    int fact = 1;
    for (int i = 2; i < n; i++) {
      fact = _mul(fact, i);
    }
    fact = _rev(fact);
    for (int i = n; i >= 1; i--) {
      C[i] = _mul(C[i], fact);
      fact = _mul(fact, i - 1);
    }
  }
  dfs(1);
  vector<int> p = f[1].gen_polynom().a;
  p.erase(p.begin());
  p.resize(n + 1);
  int res = 0;
  for (int i = 1; i <= n; i++) {
    if ((res += _mul(C[i], p[i])) >= mod) {
      res -= mod;
    }
  }
  printf("%d\n", res);
  return 0;
}