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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

def __logging(visited, rest=[]):
    if rest:
        print(' visited:'+str(visited)+'\n'+
              '    rest:'+str(rest)+'\n')
    else:
        print(' visited:'+str(visited))


def dfs(graph, start, visited):
    if start in visited:
        return True
    
    stack = [(start, 0)]
    distance = []
    flag = True
    while stack:
        temp = stack.pop(0)
        label = temp[0]
        dis = temp[1]
        if label not in visited:
            tmp = graph.get(label, [])
            tmp = list(map(lambda x: (x[0], x[1] + dis), tmp))
            stack = tmp + stack
            visited[label] = dis
        else:
            if flag and dis != visited[label]:
                flag = False
        #__logging(visited, stack)
    return flag

N, M = (int(i) for i in input().split())  

m_list = []
graph = {}

for i in range(M):
    L, R, D = (int(i) for i in input().split())
    if L not in graph:
        graph[L] = [(R, D)]
        m_list.append(L)
    else:
        graph[L].append((R, D))
    if R not in graph:
        graph[R] = [(L, -1 * D)]
        m_list.append(R)
    else:
        graph[R].append((L, -1 * D))

visited = {}
flag = True
for i in range(1, N + 1):
    ans = dfs(graph, i, visited)
    if not ans:
        flag = False
        break

if flag:
    print('Yes')
else:
    print('No')