Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

N,M = map(int,input().split())
info = [list(map(int,input().split())) for i in range(M)]

def root(x):
    # ノードが負 <=> xは根
    if nodes[x] < 0:
        return x
    else:
        # 経路圧縮:再帰的に根を調べてその過程で通った要素は全て根に直接つなぎ直す
        nodes[x] = root(nodes[x])
        # 親の重みの累積和をとる. これがxを根に直接繋いだ時の重みになる
        weight[x] += weight[nodes[x]]
        return nodes[x]

def unite(x,y,w):
    root_x = root(x); root_y = root(y)
    # 根が同じなら結合済みなのでreturn(わざわざ分岐するのは自己ループを考慮)
    if root_x == root_y:
        return
    #根ノードは木の高さを表す(負数で表していることに注意)
    rank_x = -nodes[root_x]; rank_y = -nodes[root_y]
    if rank_x >= rank_y:
        root_l = root_x; root_s = root_y
        l = x; s = y
    else:
        root_l = root_y; root_s = root_x
        l = y; s = x
        w *= -1
    # 繋いだ分だけ木を高くする
    nodes[root_l] += nodes[root_s]
    # 低い木を高い木に繋げる
    nodes[root_s] = root_l
    # 繋ぎ直す時の重みの変更
    weight[root_s] = w - weight[s] + weight[l]

# 根が同じ->True
def same(x,y):
    return root(x) == root(y)

# xからyへの重み
def diff(x,y):
    return weight[y] - weight[x]

# ノードは-1で初期化, Nは頂点数
nodes = [-1]*N
# ウェイトは0で初期化, Nは頂点数
weight = [0]*N
ans = "Yes"
for L,R,D in info:
    L -= 1; R -= 1
    if same(L,R) is True and diff(L,R) != D:
        ans = "No"
        break
    unite(L,R,D)
print(ans)