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())

adjList = [[] for i in range(N)]
for i in range(M):
    L, R, D = map(int, input().split())
    adjList[L - 1].append((R - 1, D))
    adjList[R - 1].append((L - 1, -D))

def bfs(vSt):
    xs[vSt] = 0
    Q = deque()
    Q.append(vSt)
    while Q:
        vNow = Q.popleft()

        for v2, d in adjList[vNow]:
            x2 = xs[vNow] + d
            if xs[v2] == None:
                xs[v2] = x2
                Q.append(v2)
            elif xs[v2] != x2:
                return False

    return True


xs = [None] * N
for vSt in range(N):
    if xs[vSt] == None:
        if not bfs(vSt):
            print('No')
            exit()

print('Yes')