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:
N,M = map(int,input().split())
A = [[] for i in range(N+1)]

for i in range(M) :
    l,r,d = map(int,input().split())
    
    l -= 1
    r -= 1
    
    A[l].append([r,d])
    A[r].append([l,-d])

distance = ["#" for i in range(N)]

ans = "Yes"

flg = 0

for i in range(N) :
    if flg == 1 :
        break
    
    if distance[i] != "#" :
        continue
    
    distance[i] = 0
    stack = [i]
    
    while stack :
        v = stack.pop()
        
        for to,d in A[v] :
            if distance[to] == "#" :
                distance[to] = distance[v] + d
                stack.append(to)
            
            elif distance[to] != distance[v] + d :
                ans = "No"
                stack = []
                flg = 1
                break
                

print(ans)