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:
from collections import deque

n,m = map(int, input().split())
edges = [{} for i in range(n+1)]

for i in range(m):
    l, r, d = map(int, input().split())
    edges[l][r] = d
    edges[r][l] = d * -1

remains = set(range(1,n+1))
deq = deque()

min_ = -1 * 10**10
xs = [min_] * (n+1)

while( len(remains) > 0):
    deq.append( remains.pop() )
    while(len(deq) > 0):
        tmp = deq.popleft()
        if(xs[tmp]  == min_):
            xs[tmp] = 0
        tmp_x = xs[tmp]

        edge = edges[tmp]

        for key,val in edge.items():
            if(xs[key] == min_):
                xs[key] = tmp_x + val
                deq.append( key )
                remains.remove( key )
            else:
                if( xs[key] != tmp_x + val ):
                    print('No')
                    exit()

print('Yes')