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)

from collections import defaultdict
from heapq import heappush, heappop

def dijkstra(graph:list, node:int, start:int) -> list:
    # graph[node] = [(cost, to)]
    inf = float('inf')
    dist = [inf] * node

    dist[start] = 0
    heap = [(0, start)]
    while heap:
        cost, thisNode = heappop(heap)
        for NextCost, NextNode in graph[thisNode]:
            dist_cand = dist[thisNode] + NextCost
            if dist_cand < dist[NextNode]:
                dist[NextNode] = dist_cand
                heappush(heap,(dist[NextNode], NextNode))
    return dist
    # dist = [costs to nodes]

from collections import deque
def topological_sort(graph: list, n_v: int) -> list:
    res = []
    indegree = [0] * n_v

    for i in range(n_v):
        for c, v in graph[i]:
            indegree[v] += 1

    cand = deque()
    for i in range(n_v):
        if indegree[i] == 0:
            cand.append(i)

    while cand:
        v1 = cand.popleft()
        res.append(v1)
        for c, v2 in graph[v1]:
            indegree[v2] -= 1
            if indegree[v2] == 0:
                cand.append(v2)

    return res


def is_dag(graph: list, n_v: int):
    ts = topological_sort(graph, n_v)
    return len(ts) == n_v


n, m = map(int, input().split())
graph = defaultdict(list)
graph2 = defaultdict(list)
edges = []
l_nodes = set()
r_nodes = set()
for i in range(m):
    l, r, d = map(int, input().split())
    l -= 1
    r -= 1
    edges.append((l, r, d))
    graph[l].append((d, r))
    graph2[r].append((d, l))

    l_nodes.add(l)
    r_nodes.add(r)


if m == 0:
    print("Yes")
elif not is_dag(graph, n):
    print("No")
else:
    x = l_nodes - r_nodes

    dist = dijkstra(graph, n, list(x)[0])
    possible = True
    for l, r, d in edges:
        if dist[l] != float('inf') and dist[r] != float('inf') and d != abs(dist[l] - dist[r]):
            possible = False
            break

    if possible:
        x = r_nodes - l_nodes

        dist = dijkstra(graph2, n, list(x)[0])
        possible = True
        for l, r, d in edges:
            if dist[l] != float('inf') and dist[r] != float('inf') and d != abs(dist[l] - dist[r]):
                possible = False
                break
        if possible:
            print("Yes")
        else:
            print("No")
    else:
        print("No")