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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.height = [0] * size
        self.size = [1] * size
        self.component = size

    def root(self, index):
        if self.parent[index] == index:  # 根の場合
            return index
        rootIndex = self.root(self.parent[index])  # 葉の場合親の根を取得
        self.parent[index] = rootIndex  # 親の付け直し
        return rootIndex

    def union(self, index1, index2):  # 結合
        root1 = self.root(index1)
        root2 = self.root(index2)

        if root1 == root2:  # 連結されている場合
            return

        self.component -= 1  # 連結成分を減らす

        if self.height[root1] < self.height[root2]:
            self.parent[root1] = root2  # root2に結合
            self.size[root2] += self.size[root1]
        else:
            self.parent[root2] = root1  # root1に結合
            self.size[root1] += self.size[root2]
            if self.height[root1] == self.height[root2]:
                self.height[root1] += 1
        return

    def isSameRoot(self, index1, index2):
        return self.root(index1) == self.root(index2)

    def sizeOfSameRoot(self, index):
        return self.size[self.root(index)]

    def getComponent(self):
        return self.component

N, M = map(int, input().split())
INF = 10**18
tree = UnionFind(N)

edges = [[] for _ in range(N)]
degree = [0] * N
for _ in range(M):
    fr, to, d = map(int, input().split())
    fr -= 1
    to -= 1
    edges[to].append((fr, d))
    degree[fr] += 1
    tree.union(fr, to)

D = [INF] * N
for i in range(N):
    D[tree.root(i)] = min(D[tree.root(i)], degree[i])

if max([D[i] for i in range(N) if tree.root(i) == i]) != 0:
    print('No')
    exit()

V = set()
minDist = [INF] * N
for start in range(N):
    if degree[start] > 0:
        continue
    st = [(start, 0)]
    W = set()
    while st:
        now, dist = st.pop()
        W.add(now)

        if minDist[now] == dist:
            continue

        if minDist[now] != INF and minDist[now] != dist:
            print('No')
            exit()

        minDist[now] = dist
        for to, d in edges[now]:
            if minDist[to] == d + dist or to in V:
                continue
            st.append((to, dist + d))
    V |= W

print('Yes')