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)

import sys

sys.setrecursionlimit(10000)
INF = float('inf')


class WeightedUnionFind:
    def __init__(self, nodes):
        """
        :param collections.Iterable nodes:
        """
        self._parents = {k: k for k in nodes}
        self._ranks = {k: 0 for k in nodes}
        self._sizes = {k: 1 for k in nodes}
        # 直近の親からの重み。root なら 0
        self._weights = {k: 0 for k in nodes}

    def unite(self, x, y, w):
        """
        x が属する木と y が属する木を併合
        :param x:
        :param y:
        :param w: x と y の重みの差; (重み y) - (重み x)
        :return:
        """
        # それぞれの root をくっつけたいので root 間の重みに変換
        # x が上、y が下
        w += self.weight(x)
        w -= self.weight(y)

        x = self.find(x)
        y = self.find(y)
        if x == y:
            return

        # rank が小さい方が下
        if self._ranks[x] > self._ranks[y]:
            # x が root
            self._parents[y] = x
            self._sizes[x] += self._sizes[y]
            self._weights[x] = 0
            self._weights[y] = w
        else:
            # y が root
            self._parents[x] = y
            self._sizes[y] += self._sizes[x]
            self._weights[x] = -w
            self._weights[y] = 0
            if self._ranks[x] == self._ranks[y]:
                self._ranks[y] += 1

    def find(self, x):
        """
        x が属する木の root
        :param x:
        :return:
        """
        if self._parents[x] == x:
            return x
        root = self.find(self._parents[x])
        self._weights[x] += self._weights[self._parents[x]]
        self._parents[x] = root
        return root

    def size(self, x):
        """
        x が属する木のノード数
        :param x:
        :return:
        """
        return self._sizes[self.find(x)]

    def weight(self, x):
        """
        :param x:
        :return:
        """
        # 経路圧縮
        self.find(x)
        return self._weights[x]

    def diff(self, x, y):
        """
        (y の重み) - (x の重み)
        :param x:
        :param y:
        :return:
        """
        if self.find(x) == self.find(y):
            return self.weight(y) - self.weight(x)
        return float('inf')


N, M = list(map(int, input().split()))

uf = WeightedUnionFind([i + 1 for i in range(N)])
ok = True
for _ in range(M):
    l, r, d = list(map(int, input().split()))
    if uf.find(l) == uf.find(r) and uf.diff(l, r) != d:
        ok = False
        break
    uf.unite(l, r, d)

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