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:
from collections import*
class UnionFind():
    def __init__(self,size):
        self.table=[-1]*size
    def root(self,x):
        while self.table[x]>=0:
            x=self.table[x]
        return x
    def unite(self,x,y):
        s1=self.root(x)
        s2=self.root(y)
        if s1!=s2:
            if self.table[s1]>self.table[s2]:
                s1,s2=s2,s1
            self.table[s1]+=self.table[s2]
            self.table[s2]=s1
        return
    def same(self,x,y):
        return self.root(x)==self.root(y)
    def size(self,x):
        return -self.table[self.root(x)]
n,m,*t=map(int,open(0).read().split())
u=UnionFind(n+1)
f=defaultdict(list)
e=defaultdict(list)
for l,r,d in zip(t[::3],t[1::3],t[2::3]):
    u.unite(l,r)
    e[l].append((r,d))
for i in range(n+1):
    f[u.root(i)].append(i)
for t in f.values():
    if len(t)<3:continue
    outs=defaultdict(list)
    ins=defaultdict(int)
    for i in t:
        outs[i]=[j for j,v in e[i]]
        for j in outs[i]:
            ins[j]+=1
    q=deque(v1 for v1 in t if ins[v1]==0)
    res=[]
    while q:
        v1=q.popleft()
        res.append(v1)
        for v2 in outs[v1]:
            ins[v2]-=1
            if ins[v2]==0:
                q.append(v2)
    if not res:
        print('No')
        exit()
    c=defaultdict(int)
    for i in res:
        for j,v in e[i]:
            if c[j]and c[j]!=c[i]+v:
                print('No')
                exit()
            c[j]=c[i]+v
print('Yes')