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: Wrong Answer)

#PyPy遅え
import sys
sys.setrecursionlimit(100000000)
inf=1000000000000000000
n,m=map(int,input().split())
x=[inf]*n
#まだ訪問してなければTrue
check=[True]*n
#どこと関係があるかを記録する(maybe right)
lrd=[[] for i in range(n)]
for i in range(m):
    l,r,d=map(int,input().split())
    l-=1
    r-=1
    lrd[l].append((r,d))
    lrd[r].append((l,-d))
#wrong thought
#どっちも決まってない場合の処理
#決まってるとこからやれば良い
#ソートして順番に決めておく

#記録した関係から辿っていく
#まず初めのやつ
st=-1
for i in range(n):
    if len(lrd[i])>=1:
        st=i
        break

def bfs(z):
    #globalくらいちゃんとしろ
    global lrd,check,x
    for i in range(len(lrd[z])):
        k=lrd[z][i]
        if check[k[0]]:
            x[k[0]]=x[z]+k[1]
            check[k[0]]=False
            bfs(k[0])
if st==-1:
    print("Yes")
    sys.exit()
x[st]=0
check[st]=False
bfs(st)

for i in range(n):
    l=len(lrd[i])
    for j in range(l):
        k=lrd[i][j]
        if x[k[0]]!=x[i]+k[1]:
            print("No")
            sys.exit()
print("Yes")