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:
class WeighttedUnionFind:
	def __init__(self, n):
		self.par = [i for i in range(n+1)]
		self.rank = [0] * (n+1)
		#根への距離を管理
		self.weight = [0] * (n+1)

	#検索
	def find(self, x):
		if self.par[x] ==x:
			return x
		else:
			y = self.find(self.par[x])
			#根への重みを追加しながら根まで走査
			self.weight[x] +=self.weight[self.par[x]]
			self.par[x] = y
			return y

	def union(self, x, y, w):
		rx = self.find(x)
		ry = self.find(y)

		#xの木の高さ < yの木の高さ
		if self.rank[rx] < self.rank[ry]:
			self.par[rx] = ry
			self.weight[rx] = w - self.weight[x] + self.weight[y]

		#xの木の高さ>=yの木の高さ
		else:
			self.par[ry] = rx
			self.weight[ry] = -w - self.weight[y] + self.weight[x]
			#木の高さが同じだった場合の処理
			if self.rank[rx] == self.rank[ry]:
				self.rank[rx] +=1
	#同じ集合に属するか
	def same(self, x, y):
		return self.find(x) == self.find(y)
	#xからyへのコスト
	def diff(self, x, y):
		return self.weight[x] - self.weight[y]

N, M = map(int, input().split())
wuf = WeighttedUnionFind(N)

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

	if not wuf.same(l, r):
		wuf.union(l, r, d)

	else:
		if wuf.diff(l, r) !=d:
			print("No")
			exit()
print("Yes")