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 C (GCC 5.4.1) (judge verdict: Accepted)

/* Coached by Dukkha */
#include <stdio.h>

#define N 100000
#define M 200000
#define INF 0x3f3f3f3f

int oo[1 + M * 2], oj[1 + M * 2], ow[1 + M * 2];

int link(int o, int j, int w) {
	static int _ = 1;

	oo[_] = o, oj[_] = j, ow[_] = w;
	return _++;
}

int ae[N], xx[N];

int dfs(int p, int i, int x) {
	int o;

	if (xx[i])
		return xx[i] == x;
	xx[i] = x;
	for (o = ae[i]; o; o = oo[o]) {
		int j = oj[o], w = ow[o];

		if (j != p && !dfs(i, j, x + w))
			return 0;
	}
	return 1;
}

int main() {
	int n, m, i, j;

	scanf("%d%d", &n, &m);
	while (m--) {
		int w;

		scanf("%d%d%d", &i, &j, &w), i--, j--;
		ae[i] = link(ae[i], j, w);
		ae[j] = link(ae[j], i, -w);
	}
	for (i = 0; i < n; i++)
		if (xx[i] == 0 && !dfs(-1, i, INF)) {
			printf("No\n");
			return 0;
		}
	printf("Yes\n");
	return 0;
}