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;
int n, q;
vector<int> v[150007];
int subtree[150007];
int heavy[150007];
int lst[150007];
int root[150007];
int tpsz;
int ord[150007];
int st[150007];
int en[150007];
long long add[150007];
void dfs(int vertex, int prv) {
  ord[++tpsz] = vertex;
  st[vertex] = tpsz;
  subtree[vertex] = 1;
  int sz = v[vertex].size();
  for (int i = 0; i < sz; ++i) {
    if (v[vertex][i] == prv) {
      continue;
    }
    lst[v[vertex][i]] = vertex;
    dfs(v[vertex][i], vertex);
    subtree[vertex] += subtree[v[vertex][i]];
    if (subtree[v[vertex][i]] > subtree[heavy[vertex]]) {
      heavy[vertex] = v[vertex][i];
    }
  }
  en[vertex] = tpsz;
}
long long fastpow(long long x, long long st) {
  long long ret = 1;
  while (st > 0) {
    if ((st & 1) != 0) {
      ret = (ret * x) % 998244353;
      --st;
    } else {
      x = (x * x) % 998244353;
      st >>= 1;
    }
  }
  return ret;
}
class Tree {
 public:
  long long tr[5 * 150007];
  void init(int where, int IL, int IR) {
    tr[where] = 0;
    if (IL == IR) {
      return;
    }
    int mid = (IL + IR) / 2;
    init(2 * where, IL, mid);
    init(2 * where + 1, mid + 1, IR);
  }
  void update(int where, int IL, int IR, int CURL, int CURR, long long val) {
    if (CURL > CURR) {
      return;
    }
    if (IR < CURL || CURR < IL) {
      return;
    }
    if (CURL <= IL && IR <= CURR) {
      tr[where] += val;
      if (tr[where] >= 998244353) {
        tr[where] -= 998244353;
      }
      return;
    }
    int mid = (IL + IR) / 2;
    update(2 * where, IL, mid, CURL, CURR, val);
    update(2 * where + 1, mid + 1, IR, CURL, CURR, val);
  }
  long long query(int where, int IL, int IR, int pos) {
    if (IR < pos || pos < IL) {
      return 0;
    }
    if (IL == IR) {
      return tr[where];
    }
    long long ret = tr[where];
    int mid = (IL + IR) / 2;
    if (pos <= mid) {
      ret += query(2 * where, IL, mid, pos);
    } else {
      ret += query(2 * where + 1, mid + 1, IR, pos);
    }
    return ret;
  }
};
Tree w;
void input() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y), v[y].push_back(x);
  }
}
void solve() {
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) {
    if (heavy[lst[i]] != i) {
      int x = i;
      while (x > 0) {
        root[x] = i;
        x = heavy[x];
      }
    }
  }
  long long coef = fastpow(n, 998244353 - 2);
  w.init(1, 1, n);
  while (q--) {
    int type, x, y;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d", &x, &y);
      add[x] += y;
      if (add[x] >= 998244353) {
        add[x] -= 998244353;
      }
      long long aux = (1LL * subtree[x] * y) % 998244353;
      w.update(1, 1, n, 1, st[x] - 1, aux);
      w.update(1, 1, n, en[x] + 1, n, aux);
      if (heavy[x] > 0) {
        aux = (1LL * (n - subtree[heavy[x]]) * y) % 998244353;
        w.update(1, 1, n, st[heavy[x]], en[heavy[x]], aux);
      }
    } else {
      scanf("%d", &x);
      long long ans = 1LL * n * add[x] + w.query(1, 1, n, st[x]);
      ans %= 998244353;
      while (x > 0) {
        x = root[x];
        ans += (1LL * add[lst[x]] * (n - subtree[x])) % 998244353;
        if (ans >= 998244353) {
          ans -= 998244353;
        }
        x = lst[x];
      }
      printf("%I64d\n", (ans * coef) % 998244353);
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}