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: Wrong Answer)

import sys

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline


class WeightedUnionFind:
    def __init__(self, N):
        self.par = [i for i in range(N)]
        self.rank = [0] * N
        self.weight = [0] * N  # 親までの距離

    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            r = self.find(self.par[x])
            self.weight[x] += self.weight[self.par[x]]
            self.par[x] = r
            return r

    def union(self, x, y, w):
        rx = self.find(x)
        ry = self.find(y)
        if self.rank[rx] < self.rank[ry]:
            self.par[rx] = ry
            self.weight[rx] = -w - self.weight[x] + self.weight[y]
        else:
            self.par[ry] = rx
            self.weight[ry] = w - self.weight[y] + self.weight[x]
            if self.rank[rx] == self.rank[ry]:
                self.rank[rx] += 1

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def diff_w(self, x, y):
        return self.weight[x] - self.weight[y]


def main():
    N, M = map(int, readline().split())
    m = map(int, read().split())
    LRD = list(zip(m, m, m))

    D = WeightedUnionFind(N)

    for l, r, d in LRD:
        l -= 1
        r -= 1
        D.union(l, r, d)

    for i in D.par:
        if D.weight[i] != 0:
            print('No')
            sys.exit()

    print('Yes')


main()