Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		UnionFind uf = new UnionFind(n);
		boolean ok=true;
		for(int i=0;i<m;i++) {
			ok = ok && uf.union(sc.nextInt()-1, sc.nextInt()-1, sc.nextInt());
		}
		if(ok) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}
}

class UnionFind {
	int[] parent;
	int[] rank;
	int[] distanceFromParent;
	UnionFind(int num) {
		this.parent=new int[num];
		for(int i=0;i<num;i++) {
			this.parent[i]=i;
		}
		this.rank=new int[num];
		this.distanceFromParent=new int[num];
	}

	public int find(int num) {
		if(this.parent[num]==num) {
			return num;
		} else {
			return find(this.parent[num]);
		}
	}

	public int distanceFromRoot(int num) {
		if(this.parent[num]==num) {
			return 0;
		} else {
			return distanceFromRoot(this.parent[num])+distanceFromParent[num];
		}
	}

	public boolean union(int l, int r, int d) {
		int lRoot=find(l);
		int rRoot=find(r);

		if(lRoot==rRoot) {
			//check consistency
			if(distanceFromRoot(r)-distanceFromRoot(l)==d) {
				return true;
			} else {
				return false;
			}
		} else if(rank[lRoot]<rank[rRoot]) {
			parent[lRoot]=rRoot;
			distanceFromParent[lRoot]=-d-distanceFromRoot(l)+distanceFromRoot(r);
		} else if(rank[lRoot]>rank[rRoot]) {
			parent[rRoot]=lRoot;
			distanceFromParent[rRoot]=d-distanceFromRoot(r)+distanceFromRoot(l);
		} else {
			parent[rRoot]=lRoot;
			distanceFromParent[rRoot]=d-distanceFromRoot(r)+distanceFromRoot(l);
			rank[lRoot]++;
		}
		return true;
	}

}