Problem: Hanh is a famous biologist. He loves growing trees and doing experiments on his own garden.

One day, he got a tree consisting of n vertices. Vertices are numbered from 1 to n. A tree with n vertices is an undirected connected graph with n-1 edges. Initially, Hanh sets the value of every vertex to 0.

Now, Hanh performs q operations, each is either of the following types: 

  * Type 1: Hanh selects a vertex v and an integer d. Then he chooses some vertex r uniformly at random, lists all vertices u such that the path from r to u passes through v. Hanh then increases the value of all such vertices u by d. 
  * Type 2: Hanh selects a vertex v and calculates the expected value of v. 



Since Hanh is good at biology but not math, he needs your help on these operations.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 150 000) — the number of vertices on Hanh's tree and the number of operations he performs.

Each of the next n - 1 lines contains two integers u and v (1 ≤ u, v ≤ n), denoting that there is an edge connecting two vertices u and v. It is guaranteed that these n - 1 edges form a tree.

Each of the last q lines describes an operation in either formats: 

  * 1 v d (1 ≤ v ≤ n, 0 ≤ d ≤ 10^7), representing a first-type operation. 
  * 2 v (1 ≤ v ≤ n), representing a second-type operation. 



It is guaranteed that there is at least one query of the second type.

Output

For each operation of the second type, write the expected value on a single line. 

Let M = 998244353, it can be shown that the expected value can be expressed as an irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output the integer equal to p ⋅ q^{-1} mod M. In other words, output such an integer x that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Example

Input


5 12
1 2
1 3
2 4
2 5
1 1 1
2 1
2 2
2 3
2 4
2 5
1 2 2
2 1
2 2
2 3
2 4
2 5


Output


1
199648871
399297742
199648871
199648871
598946614
199648873
2
2
2

Note

The image below shows the tree in the example:

<image>

For the first query, where v = 1 and d = 1: 

  * If r = 1, the values of all vertices get increased. 
  * If r = 2, the values of vertices 1 and 3 get increased. 
  * If r = 3, the values of vertices 1, 2, 4 and 5 get increased. 
  * If r = 4, the values of vertices 1 and 3 get increased. 
  * If r = 5, the values of vertices 1 and 3 get increased. 



Hence, the expected values of all vertices after this query are (1, 0.4, 0.8, 0.4, 0.4).

For the second query, where v = 2 and d = 2: 

  * If r = 1, the values of vertices 2, 4 and 5 get increased. 
  * If r = 2, the values of all vertices get increased. 
  * If r = 3, the values of vertices 2, 4 and 5 get increased. 
  * If r = 4, the values of vertices 1, 2, 3 and 5 get increased. 
  * If r = 5, the values of vertices 1, 2, 3 and 4 get increased. 



Hence, the expected values of all vertices after this query are (2.2, 2.4, 2, 2, 2).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 1.5e5 + 10;
const long long Mod = 998244353;
long long power(long long x, long long y, long long res = 1) {
  for (; y; y >>= 1) {
    if (y & 1) {
      res = 1ll * res * x % Mod;
    }
    x = 1ll * x * x % Mod;
  }
  return res;
}
struct Edge {
  long long to, nxt;
} e[N << 1];
struct Segment_Tree {
  long long pre, tag;
} t[N << 2];
long long n, T, tot, cnt, Q;
long long head[N], dep[N], f[N], siz[N], son[N], pos[N], top[N], tag[N];
void Build(long long x, long long y) {
  e[++tot].to = y;
  e[tot].nxt = head[x];
  head[x] = tot;
  e[++tot].to = x;
  e[tot].nxt = head[y];
  head[y] = tot;
}
void dfs1(long long x, long long fa) {
  dep[x] = dep[fa] + 1;
  f[x] = fa;
  siz[x] = 1;
  for (long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].to;
    if (y == fa) {
      continue;
    }
    dfs1(y, x);
    siz[x] += siz[y];
    if (siz[y] > siz[son[x]]) {
      son[x] = y;
    }
  }
}
void dfs2(long long x, long long Top) {
  pos[x] = ++cnt;
  top[x] = Top;
  if (son[x]) {
    dfs2(son[x], Top);
  }
  for (long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].to;
    if (y == f[x] || y == son[x]) {
      continue;
    }
    dfs2(y, y);
  }
}
void spread(long long p) {
  if (!t[p].tag) {
    return;
  }
  t[p << 1].pre += t[p].tag;
  t[p << 1 | 1].pre += t[p].tag;
  t[p << 1].tag += t[p].tag;
  t[p << 1 | 1].tag += t[p].tag;
  t[p].tag = 0;
}
void Modify(long long p, long long l, long long r, long long ql, long long qr,
            long long k) {
  if (ql > qr) {
    return;
  }
  if (l >= ql && r <= qr) {
    t[p].pre += k;
    t[p].tag += k;
    return;
  }
  spread(p);
  long long mid = l + r >> 1;
  if (mid >= ql) {
    Modify(p << 1, l, mid, ql, qr, k);
  }
  if (mid < qr) {
    Modify(p << 1 | 1, mid + 1, r, ql, qr, k);
  }
}
long long query(long long p, long long l, long long r, long long k) {
  if (l == r) {
    return t[p].pre;
  }
  spread(p);
  long long mid = l + r >> 1;
  if (mid >= k) {
    return query(p << 1, l, mid, k);
  }
  return query(p << 1 | 1, mid + 1, r, k);
}
long long Range_query(long long x) {
  long long res = (1ll * n * tag[x] + query(1, 1, n, pos[x])) % Mod;
  while (top[x] != 1) {
    res = (res + tag[f[top[x]]] * (n - siz[top[x]]) % Mod) % Mod;
    x = f[top[x]];
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> T;
  Q = power(n, Mod - 2);
  for (long long i = 1; i <= n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    Build(x, y);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  while (T-- > 0) {
    long long opt, x, d;
    cin >> opt;
    if (opt == 1) {
      cin >> x >> d;
      tag[x] = (tag[x] + d) % Mod;
      if (son[x]) {
        Modify(1, 1, n, pos[son[x]], pos[son[x]] + siz[son[x]] - 1,
               1ll * d * (n - siz[son[x]]) % Mod);
      }
      Modify(1, 1, n, 1, pos[x] - 1, 1ll * d * siz[x] % Mod);
      Modify(1, 1, n, pos[x] + siz[x], n, 1ll * d * siz[x] % Mod);
    } else {
      cin >> x;
      cout << 1ll * Range_query(x) * Q % Mod << endl;
    }
  }
  return 0;
}