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)

setrecursionlimit(1000000)
N,M = list(map(int,input().split()))

edges = [[] for _ in range(N)]

for i in range(M):
    L,R,D = list(map(int,input().split()))
    edges[L - 1].append((R -1,D))
    edges[R - 1].append((L-1, -D))


visited = [None] * N
place = [0] * N
inf = 10 ** 10


def walk(i):
    if visited[i]:
        return (place[i], place[i])
    visited[i] = True
    (n,m) = (inf, - inf)
    for j,d in edges[i]:
        #print(j,d)
        if visited[j] and place[j] != place[i] + d:
            print("No")
            exit()
        place[j] = place[i] + d
        (x,y) = walk(j)
        (n,m) = (min(n,x),max(m,y))
    return (n,m)


for i in range(N):
    if not visited[i]:
        place[i] = 0
    # 連結成分のminとmax
        (n,m) = walk(i)
        if m - n > 10 ** 9:
            print("No")
            exit()

print("Yes")