Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in PYTHON3):

Solution:
import sys
def input(): return sys.stdin.readline().strip()
N,M=map(int,input().split())
P=[list(map(int,input().split())) for i in range(M)]
inf=10**15
c=[[] for i in range(N)]
for i,j,w in P:
    c[i-1].append((j-1,w))
    c[j-1].append((i-1,-w))
d=[inf]*N
v=[0]*N
a=0
import sys
sys.setrecursionlimit(10**7)
def dfs(p,v,d):
    for n,w in c[p]:
        if v[n]==0:
            d[n]=d[p]+w
            v[n]=1
            if dfs(n,v,d)==-1:
                return -1
        elif d[p]+w!=d[n]:
            return -1
    return 0
s=0
while True:
    d[s]=0
    v[s]=1
    a=dfs(s,v,d)
    if a<0:
        break
    try:
        s=d.index(inf,s)
    except ValueError:
        break
print(['Yes','No'][a])