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)

class Node:
    def __init__(self, parent, value):
        self.parent = parent
        self.value = value

class UnionFindTree:
    def __init__(self, size):
        self.table = [Node(i, 0) for i in range(size)]
    
    def hoge(self, x):
        while self.table[x].parent != self.table[self.table[x].parent].parent:
            parent = self.table[x].parent
            self.table[x].value += self.table[parent].value
            self.table[x].parent = self.table[parent].value

    def find_root(self, x):
        self.hoge(x)
        return self.table[x].parent

    def find_value(self, x):
        self.hoge(x)
        return self.table[x].value

    def union(self, x, y, d):
        r1 = self.find_root(x)
        r2 = self.find_root(y)
        v1 = self.find_value(x)
        v2 = self.find_value(y)
        print(r1, r2, v1, v2)
        if r1 != r2:
            self.table[r2].parent = r1
            self.table[r2].value = (v1 - v2) + d
        else:
            if (v2 - v1) != d:
                return False
        return True
    
    def __repr__(self):
        return repr(self.table)

def main():
    n, m = map(int, input().split())
    tree = UnionFindTree(n+1)
    for i in range(m):
        li, ri, di = map(int, input().split())
        if not tree.union(li, ri, di):
            print('No')
            return
    print('Yes')

main()