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

n, m = map(int, input().split())
ed = [[] for _ in range(n)]
v = [None] * n
for _ in range(m):
	l, r, d = map(int, input().split())
	ed[l-1].append([r-1, d])
	ed[r-1].append([l-1, -d])

visited = [False] * n

def dfs(start):
	global ed, v, visited
	s = [start]
	v[start] = 0
	while s:
		label = s.pop()
		if visited[label] == False:
			visited[label] = True
			for e in ed[label]:
				if v[e[0]] is None:
					v[e[0]] = v[label] + e[1]
					s.append(e[0])
				else:
					if v[e[0]] != v[label] + e[1]:
						print("No")
						sys.exit()
	return

for i in range(n):
	if v[i] is None:
		dfs(i)

print("Yes")