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 Counter,defaultdict,deque
from heapq import heappop,heappush,heapify
import sys,bisect,math,itertools,fractions,pprint
sys.setrecursionlimit(10**8)
mod = 10**9+7
mod2 = 998244353
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpln(n): return list(int(sys.stdin.readline()) for i in range(n))

class WeightedUnionFind:
    def __init__(self, n):
        self.par = [i for i in range(n+1)]
        self.rank = [0] * (n+1)
        # 根への距離を管理
        self.weight = [0] * (n+1)

    # 検索
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            y = self.find(self.par[x])
            # 親への重みを追加しながら根まで走査
            self.weight[x] += self.weight[self.par[x]]
            self.par[x] = y
            return y

    # 併合
    def union(self, x, y, w):
        rx = self.find(x)
        ry = self.find(y)
        # xの木の高さ < yの木の高さ
        if self.rank[rx] < self.rank[ry]:
            self.par[rx] = ry
            self.weight[rx] = w - self.weight[x] + self.weight[y]
        # xの木の高さ ≧ yの木の高さ
        else:
            self.par[ry] = rx
            self.weight[ry] = -w - self.weight[y] + self.weight[x]
            # 木の高さが同じだった場合の処理
            if self.rank[rx] == self.rank[ry]:
                self.rank[rx] += 1

    # 同じ集合に属するか
    def same(self, x, y):
        return self.find(x) == self.find(y)

    # xからyへのコスト
    def diff(self, x, y):
        return self.weight[x] - self.weight[y]

def chk(a,b,c):
    return True if uf.diff(a,b) == c else False
n,m = inpl()
z = [-1] * n
uf = WeightedUnionFind(n)
for i in range(m):
    a,b,c = inpl()
    a -= 1; b -= 1
    if uf.same(a,b):
        if not chk(a,b,c):
            print('No')
            quit()
    else:
        uf.union(a,b,c)
print('Yes')