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 int V = 151000;
const int E = 310000;
struct Edge {
  int v, next;
} edge[E];
int head[V], e;
void erase() {
  memset(head, -1, sizeof(head));
  e = 0;
}
void increase(int u, int v) {
  edge[e].v = v;
  edge[e].next = head[u];
  head[u] = e++;
}
int dep[V], fa[V], size[V], son[V], w[V], we[V], z, top[V];
long long an[V];
void dfs1(int k) {
  size[k] = 1, son[k] = 0;
  an[k] = 0;
  for (int t = head[k]; t != -1; t = edge[t].next) {
    int v = edge[t].v;
    if (v == fa[k]) continue;
    dep[v] = dep[k] + 1;
    fa[v] = k;
    dfs1(v);
    if (size[v] > size[son[k]]) son[k] = v;
    size[k] += size[v];
  }
}
void dfs2(int k, int tp) {
  w[k] = ++z, top[k] = tp;
  if (son[k]) dfs2(son[k], tp);
  for (int t = head[k]; t != -1; t = edge[t].next) {
    int v = edge[t].v;
    if (v == son[k] || v == fa[k]) continue;
    dfs2(v, v);
  }
  we[k] = z;
}
struct ss {
  int l, r;
  long long sum, lzs;
  long long a, lza;
} bn[V * 4];
void buildtree(int k, int l, int r) {
  bn[k].l = l, bn[k].r = r;
  bn[k].sum = bn[k].lzs = 0;
  bn[k].a = bn[k].lza = 0;
  if (l == r) {
    return;
  }
  int lk = k << 1, rk = lk | 1;
  int mid = (l + r) / 2;
  buildtree(lk, l, mid);
  buildtree(rk, mid + 1, r);
}
void pushdown(int k) {
  int lk = k << 1, rk = lk | 1;
  bn[lk].lzs += bn[k].lzs;
  bn[rk].lzs += bn[k].lzs;
  bn[lk].sum += (bn[lk].r - bn[lk].l + 1) * bn[k].lzs;
  bn[rk].sum += (bn[rk].r - bn[rk].l + 1) * bn[k].lzs;
  bn[lk].lza += bn[k].lza;
  bn[rk].lza += bn[k].lza;
  bn[lk].a += bn[k].lza;
  bn[rk].a += bn[k].lza;
  bn[k].lzs = bn[k].lza = 0;
}
void adds(int k, int l, int r, long long a) {
  if (bn[k].l == l && bn[k].r == r) {
    bn[k].sum += (long long)(r - l + 1) * a;
    bn[k].lzs += a;
    return;
  }
  if (bn[k].lzs || bn[k].lza) pushdown(k);
  int lk = k << 1, rk = lk | 1;
  if (bn[lk].r >= r)
    adds(lk, l, r, a);
  else if (bn[rk].l <= l)
    adds(rk, l, r, a);
  else {
    adds(lk, l, bn[lk].r, a);
    adds(rk, bn[rk].l, r, a);
  }
  bn[k].sum = bn[lk].sum + bn[rk].sum;
}
void adda(int k, int l, int r, long long a) {
  if (bn[k].l == l && bn[k].r == r) {
    bn[k].a += a;
    bn[k].lza += a;
    return;
  }
  if (bn[k].lzs || bn[k].lza) pushdown(k);
  int lk = k << 1, rk = lk | 1;
  if (bn[lk].r >= r)
    adda(lk, l, r, a);
  else if (bn[rk].l <= l)
    adda(rk, l, r, a);
  else {
    adda(lk, l, bn[lk].r, a);
    adda(rk, bn[rk].l, r, a);
  }
  bn[k].sum = bn[lk].sum + bn[rk].sum;
}
long long querys(int k, int l, int r) {
  if (bn[k].l == l && bn[k].r == r) {
    return bn[k].sum;
  }
  if (bn[k].lzs || bn[k].lza) pushdown(k);
  int lk = k << 1, rk = lk | 1;
  long long ans = 0;
  if (bn[lk].r >= r)
    ans = querys(lk, l, r);
  else if (bn[rk].l <= l)
    ans = querys(rk, l, r);
  else
    ans = querys(lk, l, bn[lk].r) + querys(rk, bn[rk].l, r);
  bn[k].sum = bn[lk].sum + bn[rk].sum;
  return ans;
}
long long querya(int k, int id) {
  if (bn[k].l == id && bn[k].r == id) {
    return bn[k].a;
  }
  if (bn[k].lzs || bn[k].lza) pushdown(k);
  int lk = k << 1, rk = lk | 1;
  long long ans = 0;
  if (bn[lk].r >= id)
    ans = querya(lk, id);
  else if (bn[rk].l <= id)
    ans = querya(rk, id);
  bn[k].sum = bn[lk].sum + bn[rk].sum;
  return ans;
}
long long find(int u, int v) {
  int fu = top[u], fv = top[v];
  long long ans = 0;
  while (fu != fv) {
    if (dep[fu] < dep[fv]) {
      swap(u, v);
      swap(fu, fv);
    }
    ans += querys(1, w[fu], w[u]);
    if (an[fa[fu]]) ans += an[fa[fu]] * (size[1] - size[fu]);
    u = fa[fu], fu = top[u];
  }
  if (dep[u] > dep[v]) swap(u, v);
  return ans + querys(1, w[u], w[v]);
}
long long md = 998244353;
long long x, y;
void exgcd(long long a, long long b) {
  if (b == 0) {
    x = 1, y = 0;
    return;
  }
  exgcd(b, a % b);
  long long k;
  k = x;
  x = y;
  y = k - (a / b) * y;
}
int main() {
  int n, m;
  int i, j, k;
  int u, v;
  cin >> n >> m;
  exgcd(n, md);
  long long nn = (x + md) % md;
  erase();
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    increase(u, v);
    increase(v, u);
  }
  dep[1] = size[0] = 0, fa[1] = 0;
  dfs1(1);
  z = 0;
  dfs2(1, 1);
  buildtree(1, 1, n);
  for (i = 0; i < m; i++) {
    int t;
    scanf("%d", &t);
    int id, a;
    if (t == 1) {
      scanf("%d%d", &id, &a);
      if (son[id])
        adds(1, w[son[id]], w[son[id]], (long long)a * (n - size[son[id]]));
      adda(1, 1, n, (long long)a * size[id]);
      adda(1, w[id], we[id], (long long)(-a) * size[id]);
      an[id] += a;
    } else {
      scanf("%d", &id);
      long long ans = 0;
      ans = (find(id, 1) + querya(1, w[id]) + an[id] * n) % md;
      ans = ans * nn % md;
      cout << ans << endl;
    }
  }
  return 0;
}