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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <vector>
#include <queue>
#include <set>
using namespace std;

typedef pair<int, int> pii;
typedef pair<pii, int> ppi;
const int SIZE = 100005;

int N, M, L, R, D;
vector<pii> adj[SIZE];
bool visited[SIZE];
long long int pos[SIZE];
bool isEnd;

void search(int here){
	visited[here] = true;
	for(int i=0; i<adj[here].size(); ++i){
		int next = adj[here][i].first;
		int dist = adj[here][i].second;
		if(visited[next]){
			if(pos[next] != pos[here] + dist){
				isEnd = true;
			}
		}
		else{
			pos[next] = pos[here] + dist;
			search(next);
		}
	}
}

int main(){
	scanf("%d %d", &N, &M);
	for(int i=0; i<M; ++i){
		scanf("%d %d %d", &L, &R, &D);
		adj[L].push_back({R, D});
		adj[R].push_back({L, -D});
	}
	for(int i=1; i<=N; ++i){
		if(!visited[i])
			search(i);
	}
	if(isEnd)
		printf("No\n");
	else
		printf("Yes\n");
	
}