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:
import sys

sys.setrecursionlimit(10**8)

def find(x):	#xの根を返す
	global table
	global diff_weight
	
	if table[x] == x:
		return x
	else:
		root = find(table[x])
		diff_weight[x] += diff_weight[table[x]] #親の更新に伴う 重みの更新
		table[x] = root	#親の更新(根を直接親にして参照距離を短く)
		return table[x]

def union(x,y,w):	#xとyを diff(x,y)=w で繋げる
	w = w - weight(y) + weight(x) 
	x = find(x)	
	y = find(y)
	
	if x == y:
		return
	
	if rank[x] < rank[y]:
		x,y = y,x
		w = -w
	
	table[y] = x
	diff_weight[y] = w
	
	if rank[x] == rank[y]:
			rank[y] += 1

def check(x,y): #xとyか繋がっているか -> return bool
	if find(x) == find(y):
		return True
	else:
		return False

def weight(x):
	find(x)	#経路圧縮
	return diff_weight[x]
		
def diff(x,y): # weight(y)-weight(x)  (繋がってる前提)
	return (weight(y) - weight(x))

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

table = [i for i in range(N)]	#木の親 table[x] == x なら根	
rank  = [1 for i in range(N)]	#木の長さ
diff_weight = [0 for i in range(N)]


for i in range(M):
	a,b,D = (map(int,input().split()))
	a -= 1
	b -= 1
	if check(a,b):
		if diff(a,b) != D:
			print('No')
			sys.exit()
	else:
		union(a,b,D)
	
	
print ('Yes')