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:
class UnionFind:
    def __init__(self,n):
        self.par = [i for i in range(n+1)]  # 親のノード番号
        self.rank = [0]*(n+1)
        self.diff_weight = [0]*(n+1)  # 親ノードとのweightの差
    def root(self,x):  # xの根のノード番号
        if self.par[x] == x:
            return x
        else:
            r = self.root(self.par[x])
            self.diff_weight[x] += self.diff_weight[self.par[x]]
            self.par[x] = r
            return self.par[x]
    def same_check(self,x,y):  # x,yが同じグループか否か
        return self.root(x) == self.root(y)
    def weight(self,x):
        r = self.root(x)  # 経路圧縮
        return self.diff_weight[x]
    def diff(self,x,y):
        return self.weight(y) - self.weight(x)
    def unite(self,x,y,w):  # x,yの属するグループの併合
        w += self.weight(x)
        w -= self.weight(y)
        x = self.root(x)
        y = self.root(y)
        if self.rank[x] < self.rank[y]:
            x,y = y,x
            w = -w
        if self.rank[x] == self.rank[y]:
            self.rank[x] += 1
        self.par[y] = x
        self.diff_weight[y] = w

import sys

sys.setrecursionlimit(10**7)
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))  #空白あり
def S(): return sys.stdin.readline().rstrip()

N,M = map(int,S().split())
G = UnionFind(N)
for i in range(M):
    L,R,D = LI()
    if G.same_check(L,R):
        if G.diff(L,R) != D:
            print('No')
            exit()
    else:
        G.unite(L,R,D)

print('Yes')