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: Runtime Error)

import sys
sys.setrecursionlimit(10**9)
N,M = map(int, input().split())
LRD = [ [int(j) for j in input().split()] for _ in range(M)]

data = [-1] * (N+1)
rank = [0]  * (N+1)
diff_weight = [0] * (N+1)

def root(x):
    if data[x] == x:
        return x
    else:
        r = root(x)
        diff_weight[x] += diff_weight[ data[x] ]
        data[x] = r
        return r

def weight(x):
    _ = root(x)
    return diff_weight[x]


def diff(x,y):
    return weight(y) - weight(x)

def merge(x,y,w):
    w  += weight(x); w -= weight(y)
    x = root(x)
    y = root(y)
    if x == y: return False
    if rank[x] < rank[y]: x,y = y,x; w = -w
    if rank[x] == rank[y]: rank[x] += 1
    data[y] = x
    diff_weight[y] = w
    return True

def issame(x,y):
    return root(x) == root(y)

for i in range(M):
    L,R,D = LRD[i]
    if not issame(L,R):
        merge(L,R,D)
    else:
        if weight(L,R) != D:
            print("No")
            exit(0)
            
print("Yes")