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:
#!/usr/bin python3
# -*- coding: utf-8 -*-

import sys
sys.setrecursionlimit(10001000)

class UnionFindWeighted:
    def __init__(self, n):                      #初期化
        self.n = n                              # 要素数
        self.parents = [i for i in range(n)]    # 親
        self.ranks = [0] * n                    # 木の深さ
        self.sizes = [1] * n                    # グループの要素数
        self.weights = [0] * n                  # 親との重み

    def find(self, x):          #親を出力
        if self.parents[x] == x:
            return x
        else:
            p = self.find(self.parents[x])
            self.weights[x] += self.weights[self.parents[x]]
            self.parents[x] = p
            return p

    def unite(self, x, y, w):          # ユニオン
        # a[x]->a[y]  の差はw  # a[y]=a[x]+w
        rx = self.find(x)
        ry = self.find(y)
        wx = self.weight(x)
        wy = self.weight(y)
        if rx == ry: return
        if self.ranks[rx] > self.ranks[ry]:
            rx , ry = ry, rx    #ryを親にする
            wx , wy = wy, wx
            w *= -1
            print("a")
        self.parents[rx] = ry
        self.sizes[ry] += self.sizes[rx]
        self.weights[rx] = wy - wx - w
        if self.ranks[rx]==self.ranks[ry]:
            self.ranks[rx] += 1

    def same(self, x, y):       #xとyが同じグループかどうか
        return self.find(x) == self.find(y)

    def members(self, x):       #xと同じグループの要素
        root = self.find(x)
        return {i for i in range(self.n) if self.find(i) == root}

    def size(self, x):          #グループの要素数
        root = self.find(x)
        return self.sizes[root]

    def roots(self):            #親の要素一覧
        return {i for i, x in enumerate(self.parents) if i == x}

    def group_count(self):      #グループの個数
        return len(self.roots())

    def all_group_members(self):      #グループのメンバー一覧
        return {r: self.members(r) for r in self.roots()}

    def weight(self, x):       #重みの差
        _ = self.find(x)
        return self.weights[x]

    def diff(self, x, y):       #重みの差
        return self.weight(y) - self.weight(x)

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

################

n, m = map(int, input().split())
uf = UnionFindWeighted(n)
for _ in range(m):
    a, b, w = map(int,input().split())
    a -= 1
    b -= 1
    if uf.same(a, b):
        if uf.diff(a, b)!=w:
            print('No')
            exit()
    else:
        uf.unite(a, b, w)
print('Yes')