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
# Your code here!
# coding: utf-8
# Your code here!
class PotUnionFind:
    __slots__ = ["parent", "gsize", "diff_p"]
    def __init__(self, n):
        self.parent = list(range(n)) #親ノード
        self.gsize = [1]*n #グループの要素数
        self.diff_p = [0]*n #親ノードを基準としたポテンシャル
 
    def root(self, x): #root(x): xの根ノードを返す．
        while self.parent[x] != x:
            self.diff_p[x] += self.diff_p[self.parent[x]]
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x 

    def weight(self, x): #root(x): xの根ノードを返す．
        c=0
        while self.parent[x] != x:
            self.diff_p[x] += self.diff_p[self.parent[x]]
            self.parent[x] = self.parent[self.parent[x]]
            c += self.diff_p[x]
            x = self.parent[x]
        return c


    def merge(self, x, y, dxy): #ポテンシャル差p(y)-p(x)=dxyでxとyの組をまとめる
        dxy += self.weight(x) - self.weight(y) #dxyをで置き換え
        x,y = self.root(x), self.root(y)
        if x == y: return False
        if self.gsize[x] > self.gsize[y]: #rxの要素数が大きいように
            x,y,dxy = y,x,-dxy
        self.gsize[x] += self.gsize[y] #xの要素数を更新
        self.parent[y] = x #ryをrxにつなぐ
        self.diff_p[y] = dxy #ryの相対ポテンシャルを更新
        return True
 
    def issame(self, x, y): #same(x,y): xとyが同じ組ならTrue
        return self.root(x) == self.root(y)
        
    def diff(self,x,y): #diff(x,y): xを基準としたyのポテンシャルを返す 
        if self.root(x) == self.root(y):
            return self.weight(y) - self.weight(x)
        else:
            return None

    def size(self,x): #size(x): xのいるグループの要素数を返す
        return self.gsize[self.root(x)]


n,m = list(map(int,input().split()))
 
uf = PotUnionFind(n)
ans = True
for i in range(m):
    l,r,d=list(map(int,input().split()))
    l-=1;r-=1;
    if uf.issame(l,r):
        ans = d==uf.diff(l,r)
    else:
        uf.merge(l,r,d)
    if ans == False:
        break
 
if ans == False:
    print('No')
else:
    print('Yes')