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 JAVA):

Solution:
//package round601;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class D2 {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), Q = ni();
		int[] from = new int[n - 1];
		int[] to = new int[n - 1];
		for (int i = 0; i < n - 1; i++) {
			from[i] = ni() - 1;
			to[i] = ni() - 1;
		}
		int[][] g = packU(n, from, to);
		int[][] pars = parents3(g, 0);
		int[] par = pars[0], ord = pars[1], dep = pars[2];
		
		int B = (int)Math.max(1, Math.sqrt(n)/2);
		int[] bigs = new int[n/B+1];
		int[] imap = new int[n];
		Arrays.fill(imap, -1);
		int bp = 0;
		for(int i = 0;i < n;i++){
			if(g[i].length >= B){
				imap[i] = bp;
				bigs[bp++] = i;
			}
		}
		bigs = Arrays.copyOf(bigs, bp);
		
		int[] des = new int[n];
		Arrays.fill(des, 1);
		for(int i = n-1;i >= 1;i--){
			des[par[ord[i]]] += des[ord[i]];
		}
		
		int[][] necks = new int[bp][n];
		int[][] ch = parentToChildren(par);
		for(int i = 0;i < bp;i++){
			Arrays.fill(necks[i], -1);
			int x = bigs[i];
			for(int e : ch[x]){
				dfs(e, e, ch, necks[i]);
			}
		}
		
		
		int[][] rights = makeRights(g, par, 0);
		int[] right = rights[2], iord = rights[1];
		
		long[] lvals = new long[bp];
		long[] ft = new long[n+5];
		long iin = invl(n, mod);
		for(int z = 0;z < Q;z++){
			int t = ni();
			if(t == 1){
				int V = ni()-1;
				long D = ni();
				if(imap[V] == -1){
					// small
					for(int e : ch[V]){
						long v = (n - des[e]) * D;
						addFenwick(ft, right[iord[e]]+1, -v);
						addFenwick(ft, iord[e], v);
					}
					long v = (des[V]) * D;
					addFenwick(ft, 0, v);
					addFenwick(ft, iord[V], D*n-v);
					addFenwick(ft, iord[V]+1, -D*n);
					addFenwick(ft, right[iord[V]]+1, v);
				}else{
					// large
					lvals[imap[V]] += D;
					if(lvals[imap[V]] >= mod)lvals[imap[V]] -= mod;
				}
			}else{
				int V = ni()-1;
				long ans = sumFenwick(ft, iord[V]);
				for(int i = 0;i < bp;i++){
					int x = bigs[i];
					if(V == x){
						ans += n * lvals[i];
					}else if(necks[i][V] != -1){
						int neck = necks[i][V];
						ans += (n - des[neck]) * lvals[i];
					}else{
						ans += des[x] * lvals[i];
					}
				}
				out.println(ans%mod*iin%mod);
			}
		}
	}
	
	static void dfs(int cur, int v, int[][] ch, int[] color)
	{
		color[cur] = v;
		for(int e : ch[cur]){
			dfs(e, v, ch, color);
		}
	}
	
	public static int[][] parentToChildren(int[] par)
	{
		int n = par.length;
		int[] ct = new int[n];
		for(int i = 0;i < n;i++){
			if(par[i] >= 0){
				ct[par[i]]++;
			}
		}
		int[][] g = new int[n][];
		for(int i = 0;i < n;i++){
			g[i] = new int[ct[i]];
		}
		for(int i = 0;i < n;i++){
			if(par[i] >= 0){
				g[par[i]][--ct[par[i]]] = i;
			}
		}
		
		return g;
	}

	
	public static long invl(long a, long mod) {
		long b = mod;
		long p = 1, q = 0;
		while (b > 0) {
			long c = a / b;
			long d;
			d = a;
			a = b;
			b = d % b;
			d = p;
			p = q;
			q = d - c * q;
		}
		return p < 0 ? p + mod : p;
	}
	
	public static int[] sortByPreorder(int[][] g, int root){
		int n = g.length;
		int[] stack = new int[n];
		int[] ord = new int[n];
		boolean[] ved = new boolean[n];
		stack[0] = root;
		int p = 1;
		int r = 0;
		ved[root] = true;
		while(p > 0){
			int cur = stack[p-1];
			ord[r++] = cur;
			p--;
			for(int e : g[cur]){
				if(!ved[e]){
					ved[e] = true;
					stack[p++] = e;
				}
			}
		}
		return ord;
	}
	
	public static int[][] makeRights(int[][] g, int[] par, int root)
	{
		int n = g.length;
		int[] ord = sortByPreorder(g, root);
		int[] iord = new int[n];
		for(int i = 0;i < n;i++)iord[ord[i]] = i;
		
		int[] right = new int[n];
		for(int i = n-1;i >= 1;i--){
			if(right[i] == 0)right[i] = i;
			int p = iord[par[ord[i]]];
			right[p] = Math.max(right[p], right[i]);
		}
		return new int[][]{ord, iord, right};
	}

	
	public static final int mod = 998244353;
	public static final long big = 8L*mod*mod;
	
	public static long sumFenwick(long[] ft, int i)
	{
		long sum = 0;
		for(i++;i > 0;i -= i&-i){
			sum += ft[i];
			if(sum >= mod)sum -= mod;
		}
		return sum;
	}
	
	public static void addFenwick(long[] ft, int i, long v)
	{
		if(v <= -mod || v >= mod)v %= mod;
		if(v < 0)v += mod;
		if(v == 0)return;
		int n = ft.length;
		for(i++;i < n;i += i&-i){
			ft[i] += v;
			if(ft[i] >= mod)ft[i] -= mod;
		}
	}


	public static int[][] parents3(int[][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);

		int[] depth = new int[n];
		depth[0] = 0;

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int nex : g[cur]) {
				if (par[cur] != nex) {
					q[r++] = nex;
					par[nex] = cur;
					depth[nex] = depth[cur] + 1;
				}
			}
		}
		return new int[][] { par, q, depth };
	}

	static int[][] packU(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new D2().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}