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>

struct nyan {
	int next;
	int diff;
};

/* pos(uf[p].next) = pos(p) + uf[p].diff */
struct nyan uf[100001];

struct nyan find_group(int p) {
	struct nyan res;
	if (uf[p].next == 0) {
		res.next = p;
		res.diff = 0;
	} else {
		struct nyan next = find_group(uf[p].next);
		res.next = next.next;
		res.diff = next.diff + uf[p].diff;
		uf[p] = res;
	}
	return res;
}

int is_same(int a, int b) {
	struct nyan an, bn;
	an = find_group(a);
	bn = find_group(b);
	return an.next == bn.next;
}

/* return pos(b) - pos(a) */
int get_diff(int a, int b) {
	struct nyan an, bn;
	an = find_group(a);
	bn = find_group(b);
	return an.diff - bn.diff;
}

void merge(int a, int b, int diff) {
	struct nyan an, bn;
	an = find_group(a);
	bn = find_group(b);
	uf[an.next].next = bn.next;
	uf[an.next].diff = bn.diff + diff - an.diff;
}

int main(void) {
	int N, M;
	int L, R, D;
	int i;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 0; i < M; i++) {
		if (scanf("%d%d%d", &L, &R, &D) != 3) return 1;
		if (is_same(L, R)) {
			if (get_diff(L, R) != D) {
				puts("No");
				return 0;
			}
		} else {
			merge(L, R, D);
		}
	}
	puts("Yes");
	return 0;
}