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: Time Limit Exceeded)

N,M = map(int,input().split())
peak = [[] for i in range(N)]
vector = []
queue = []
amount = ["n" for k in range(N)]
for i in range(M):
    temp = [int(i) for i in input().split()]
    peak[temp[0]-1].append([temp[1],temp[2]*(-1)])
    peak[temp[1]-1].append([temp[0],temp[2]])
    vector.append(temp)
 
 
f = []
 
def check(a,c):
    if amount[a-1] == "n":
        amount[a-1] = c
        queue.append(a)
        return
    else:
        if c == amount[a-1]:
            return
        else:
            f.append(0)
            return
for i in range(N):
    if f:
        break
    if amount[i] == "n":
        amount[i] = 0
        queue.append(i+1)
        while queue:
            if f:
                break
            a = queue.pop(0)
            for j in peak[a-1]:
                x,y = j[0],j[1]
                c = amount[a-1] + y
                check(x,c)
print("No" if f else "Yes")