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:
"""
unionFind木を使う。
根に対する距離を記録していく。
"""
import sys
sys.setrecursionlimit(2000000)
input = sys.stdin.readline
class UnionFind():
    def __init__(self,n):
        self.n=n
        self.parents = [i for i in range(n+1)]
        self.distance = [0]*(n+1)

    def find(self,x):
        if self.parents[x]==x:
            return x,0
        else:
            self.parents[x],dist2=self.find(self.parents[x])
            self.distance[x] += dist2
            return self.parents[x],self.distance[x]

    def union(self,l,r,dist):
        lRoot,lDist = self.find(l)
        rRoot,rDist = self.find(r)
        #まず、位置関係に矛盾がないかを調べる
        #lRootはlよりlDist右にいる
        #rRootはrよりrDist右にいる
        #lRoot=rRootのとき、lDist-rDist != distなら矛盾。
        if lRoot==rRoot and lDist-rDist != dist:
            print("No")
            exit()

        #より左にいる根を右にいる根にマージする
        if lDist < dist+rDist:
            #lRootがrRootよりも左の場合
            self.parents[lRoot]=rRoot
            self.distance[lRoot]=(dist+rDist)-lDist
        else:
            self.parents[rRoot]=lRoot
            self.distance[rRoot]=lDist-(dist+rDist)

N,M = map(int,input().split())
tree = UnionFind(N)
for i in range(M):
    L,R,D = map(int,input().split())
    tree.union(L,R,D)

print('Yes')