Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UncheckedIOException;
import java.util.StringTokenizer;
class Main{
	public static void main(String[] args) {
		SC sc=new SC(System.in);
		int N=sc.nextInt();
		int M=sc.nextInt();
		WUnionFindTree uft=new WUnionFindTree(N);
		for(int i=0; i<N; i++) {
			int L=sc.nextInt()-1;
			int R=sc.nextInt()-1;	//0 origin
			int D=sc.nextInt();
			if(uft.same(L,R)) {	//同じグループなら矛盾していないかを確認
				if(uft.diff(L,R)!=D) {
					pl("No");
					System.exit(0);
				}
				else {
					//
				}
			}
			else {	//同じグループでないならマージする。
				uft.union(L,R,D);
				if(uft.diff(L,R)!=D) {
					pl("No");
					System.exit(0);
				}
			}
		}
		pl("Yes");
	}
	static class SC {
		private BufferedReader reader = null;
		private StringTokenizer tokenizer = null;
		public SC(InputStream in) {
			reader = new BufferedReader(new InputStreamReader(in));
		}
		public String next() {
			if (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new UncheckedIOException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static void pl(Object o) {
		System.out.println(o);
	}
	public static void pl() {
		System.out.println();
	}
	public static void p(Object o) {
		System.out.print(o);
	}
	public static class WUnionFindTree{
		int[] par; // 親の番号
		int[] ws;  // 親との重みの差
		public WUnionFindTree(int n){
			par = new int[n+1];
			ws  = new int[n+1];
			for(int i = 0; i <= n; i++){ par[i] = -1; }
		}
		public int find(int x){
			if(par[x] < 0){
				return x;
			}else{
				final int parent = find(par[x]);
				ws[x] += ws[par[x]];
				return par[x] = parent;
			}
		}
		public int weight(int x){
			find(x);
			return ws[x];
		}
		public boolean union(int x, int y, int w){ // x <-(w)- y (x + w = y)
			w += weight(x);
			w -= weight(y);
			x = find(x); y = find(y);
			if(x != y){
				if(par[y] < par[x]) {
					int tmp = x; x = y; y = tmp; w = -w;
				}
				par[x] += par[y]; par[y] = x;
				ws[y] = w;
				return true;
			}else{
				return false;
			}
		}
		public boolean same(int x, int y){
			return find(x) == find(y);
		}
		public Integer diff(int x, int y){ // y - x を求める. 比較不能ならnull.
			if(!same(x, y)){ return null; }
			return this.weight(y) - this.weight(x);
		}
		public int size(int x){
			return -par[find(x)];
		}
	}
}