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)

class UnionFind:
    def __init__(self, size):
        # 負の値はルート (集合の代表) で集合の個数
        # 正の値は次の要素を表す
        self.table = [-1 for _ in range(size + 1)]

    # 集合の代表を求める
    def find(self, x):
        if self.table[x] < 0:
            return x
        else:
            # 経路の圧縮
            self.table[x] = self.find(self.table[x])
            return self.table[x]

    # 併合
    def union(self, x, y):
        s1 = self.find(x)
        s2 = self.find(y)
        if s1 != s2:
            if self.table[s1] <= self.table[s2]:
                # 小さいほうが個数が多い
                self.table[s1] += self.table[s2]
                self.table[s2] = s1
            else:
                self.table[s2] += self.table[s1]
                self.table[s1] = s2
            return True
        return False

def solve(N, M, lrd):
    uf = UnionFind(N + 1)
    for l, r, d in lrd:
        uf.union(l, r)
    xs = {}
    for i in range(1, N + 1):
        n = uf.find(i)
        if not n in xs: xs[n] = {}
        xs[n][i] = None
    for l, r, d in lrd:
        for row in xs.values():
            if l in row:
                break
        if row[l] is None:
            if row[r] is None:
                row[r] = 0
            row[l] = row[r] - d
        else:
            if row[r] is None:
                row[r] = row[l] + d
            else:
                if row[l] != row[r] - d:
                    return False
    for row in xs.values():
        vs = list(row.values())
        if vs[0] is None:
            continue
        #if max(vs) - min(vs) > 10**9:
        #   return False
    return True

if __name__ == "__main__":
    N, M = [int(x) for x in input().split()]
    lrd = []
    for i in range(M):
        lrd.append([int(x) for x in input().split()])
    if solve(N, M, lrd):
        print("Yes")
    else:
        print("No")