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())
lrd=[list(map(int,input().split())) for i in range(m)]
x=["a"]*(n+1)
#iより右にいるデータ
R=[[] for i in range(n+1)]
L=[[] for i in range(n+1)]
for u in lrd:
    l,r,d=u
    R[l].append([r,d])
    L[r].append([l,d])
for i in range(1,n+1):
    if x[i]=="a":
        x[i]=0
        que=[i]
        while que:
            h=que.pop()
            for u in R[h]:
                r,d=u
                if x[r]=="a":
                    x[r]=x[h]+d
                    que.append(r)
                else:
                    if x[r]==x[h]+d:
                        continue
                    else:
                        print("No")
                        sys.exit()
            for u in L[h]:
                l,d=u
                if x[l]=="a":
                    x[l]=x[h]-d
                    que.append(l)
                else:
                    if x[l]==x[h]-d:
                        continue
                    else:
                        print("No")
                        sys.exit()
print("Yes")