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())

# union-find改良
parent = list(range(N+1))
rank = [0] * (N+1)
pardist = [0] * (N+1) # ルートとの距離

def getRoot(x):
  if parent[x] == x:
    return x, 0
  else:
    # ルートに繋ぎ変える
    parent[x], dist = getRoot(parent[x])
    # ルートとの距離に更新
    pardist[x] += dist
    return parent[x], pardist[x]

def unite(left, right, dist):
  lRoot, lPardist = getRoot(left)
  rRoot, rPardist = getRoot(right)
  # print(parent)
  # print(pardist)
  if lRoot == rRoot:
    # 矛盾がないかチェック
    if -(rPardist - lPardist) != dist:
      print('No')
      exit()
  else:
    # 矛盾は起きない
    # ルート間の距離だけ更新して併合
    if rank[lRoot] < rank[rRoot]:
      parent[lRoot] = rRoot
      pardist[lRoot] = dist + rPardist - lPardist
    else:
      parent[rRoot] = lRoot
      pardist[rRoot] = -(dist + rPardist - lPardist)
      if rank[lRoot] == rank[rRoot]:
        rank[lRoot] += 1

for i in range(M):
  l, r, d = map(int, input().split())
  unite(l, r, d)

print('Yes')