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)

#include <stdio.h>
#include <stdlib.h>

typedef struct List {
	struct List *next;
	int v, len;
} list;

int main()
{
	int i, N, M, L[200001], R[200001], D[200001];
	scanf("%d %d", &N, &M);
	for (i = 1; i <= M; i++) scanf("%d %d %d", &(L[i]), &(R[i]), &(D[i]));
	
	list **adj = (list**)malloc(sizeof(list*) * (N + 1)), *d = (list*)malloc(sizeof(list) * M * 2);
	for (i = 0; i < M; i++) {
		d[i*2].v = R[i+1];
		d[i*2+1].v = L[i+1];
		d[i*2].len = D[i+1];
		d[i*2+1].len = -D[i+1];
		d[i*2].next = adj[L[i+1]];
		d[i*2+1].next = adj[R[i+1]];
		adj[L[i+1]] = &(d[i*2]);
		adj[R[i+1]] = &(d[i*2+1]);
	}
	
	int u, w, flag[100001] = {}, q[100001], head, tail;
	long long dist[100001];
	list* p;
	for (i = 1; i <= N; i++) {
		if (flag[i] == 1) continue;
		flag[i] = 1;
		dist[i] = 0;
		q[0] = i;
		for (head = 0, tail = 1; head < tail; head++) {
			u = q[head];
			for (p = adj[u]; p != NULL; p = p->next) {
				w = p->v;
				if (flag[w] == 0) {
					flag[w] = 1;
					dist[w] = dist[u] + p->len;
					q[tail++] = w;
				} else if (dist[w] != dist[u] + p->len) break;
			}
			if (p != NULL) break;
		}
		if (head < tail) break;
	}
	if (i > N) printf("Yes\n");
	else printf("No\n");
	fflush(stdout);
	return 0;
}