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)

#!/usr/bin/env python3
import sys
sys.setrecursionlimit(300000)

YES = "Yes"  # type: str
NO = "No"  # type: str

def solve(N: int, M: int, L: "List[int]", R: "List[int]", D: "List[int]"):
    cons = [[] for _ in range(N + 1)]
    for j in range(M):
        cons[R[j]].append([L[j], -D[j], False, len(cons[L[j]])])
        cons[L[j]].append([R[j], D[j], False, len(cons[R[j]]) - 1])

    vals = [None] * (N + 1)
    visited = [False] * (N + 1)
    #print(cons)

    def dfs(idx):
        #print(idx)
        #print(visited)
        #print(vals)
        #print()
        #visited[idx] = True
        #for t, d, used in cons[idx]:
        for con in cons[idx]:
            t = con[0]
            d = con[1]
            used = con[2]
            if not used:
                #print(idx, t, d, vals[idx], vals[t])
                if vals[idx] is None:
                    vals[idx] = 0
                if vals[t] is None:
                    vals[t] = vals[idx] + d
                elif vals[t] != vals[idx] + d:
                    return False
                con[2] = True
                cons[t][con[3]][2] = True
                if not visited[t]:
                    dfs(t)
        return True

    ret = YES
    for i in range(1, N + 1):
        if not visited[i]:
            res = dfs(i)
            if not res:
                ret = NO
                break
    #print(cons)
    print(ret)
    return


def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    M = int(next(tokens))  # type: int
    L = [int()] * (M)  # type: "List[int]" 
    R = [int()] * (M)  # type: "List[int]" 
    D = [int()] * (M)  # type: "List[int]" 
    for i in range(M):
        L[i] = int(next(tokens))
        R[i] = int(next(tokens))
        D[i] = int(next(tokens))
    solve(N, M, L, R, D)

if __name__ == '__main__':
    main()