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 int[] uf;
	public static long[] dis;
	public static int[] dep;

	public static int Root(int n) {
		if(uf[n] == n)
			return n;
		else
			return Root(uf[n]);
	}

	public static long RootDis(int n) {
		if(uf[n] == n)
			return 0;
		else
			return dis[n] + RootDis(uf[n]);
	}

	public static boolean Add(int l, int r, long d) {
		boolean ret = true;
		int lroot = Root(l);
		int rroot = Root(r);
		long tmp1 = RootDis(l);
		long tmp2 = RootDis(r);
		if(lroot != rroot) {
			if(dep[lroot] > dep[rroot]) {
				dis[rroot] = -d + tmp1 - tmp2;
				uf[rroot] = lroot;
			} else if(dep[lroot] < dep[rroot]) {
				dis[lroot] = d - tmp1 + tmp2;
				uf[lroot] = rroot;
			} else {
				dis[rroot] = -d + tmp1 - tmp2;
				uf[rroot] = lroot;
				dep[lroot]++;
			}
		} else {
			ret = (tmp1 - tmp2) == d;
		}
		return ret;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		uf = new int[n];
		dis = new long[n];
		dep = new int[n];
		for(int i = 0; i < n; i++) {
			uf[i] = i;
			dis[i] = 0;
			dep[i] = 0;
		}
		boolean ans = true;
		for(int i = 0; i < m; i++) {
			ans = ans && Add(sc.nextInt() - 1, sc.nextInt() - 1, sc.nextLong());
		}
		if(ans)
			System.out.println("Yes");
		else
			System.out.println("No");
	}

}