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:
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10000)
from collections import defaultdict, deque
def inpl(): return tuple(map(int, input().split()))
 
N, M = inpl()
Q = [inpl() for _ in range(M)]
# グループごとにトポロジカルソートをして、最後の方から遡って計算
 
tree = [[-1, 1]  for _ in range(N+1)] # [next, rank]
  
def find(i):
    if tree[i][0] == -1:
        group = i
    else:
        group = find(tree[i][0]) 
        tree[i][0] = group
    return group
  
def unite(x, y):
    px = find(x)
    py = find(y)
      
    if tree[px][1] == tree[py][1]: # rank is same
        tree[py][0] = px
        tree[px][1] += 1
    else:
        if tree[px][1] < tree[py][1]:
            px, py = py, px
        tree[py][0] = px
 
adj = [[] for _ in range(N+1)]  
 
for l, r, d in Q:
    if find(l) != find(r):
        unite(l, r)
    adj[l].append([r, d])
    adj[r].append([l, -d])
 
 
G = set([find(n) for n in range(1, N+1)])
 
X = [-1 for _ in range(N+1)]
searched = [False for _ in range(N+1)]
res = True
 
for g in G:
    X[g] = 0
    Q = deque([g])
    searched[g] = True
    while Q:
        p = Q.popleft()
        for np, d in adj[p]:
            if searched[np]:
                if X[np] != X[p] + d:
                    Q = []
                    res = False
                    break
            else:
                X[np] = X[p] + d
                Q.append(np)
                searched[np] = True
    if not res:
        break
 
if res:
    print("Yes")
else:
    print("No")