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 C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>

////////////////////////////////////////
// ポテンシャル付きUnion-Find木(未検証) //
////////////////////////////////////////

// 0-indexed
class UnionFindWithPotential {
private:
	std::vector<int> parent_, size_;
	std::vector<int64_t> potential_delta_;

	// indexの含まれる木の根のノード番号を返す
	int calcRoot(const int index)
	{
		int &parent{parent_[index]};
		if (parent == index) return index;

		parent += potential_delta_[parent];
		parent = calcRoot(parent);
		return parent;
	}
public:
	UnionFindWithPotential(const int size)
		:parent_(size), size_(size, 1), potential_delta_(size)
	{
		for (int i{}; i < size; i++) parent_[i] = i;
	}

	// index1とindex2の連結判定
	bool areConnected(const int index1, const int index2)
	{
		return calcRoot(index1) == calcRoot(index2);
	}

	// potential(index1) = potential(index2) + weightとなるよう結合
	bool unite(const int index1, const int index2, const int64_t weight)
	{
		const int root1{calcRoot(index1)}, root2{calcRoot(index2)};
		if (root1 == root2)
			return potential_delta_[index1] == potential_delta_[index2] + weight;
		if (size_[root1] <= size_[root2])
		{
			size_[root2] += size_[root1];
			potential_delta_[root1] = potential_delta_[index2] + weight - potential_delta_[index1];
			parent_[root1] = root2;
		}
		else
		{
			size_[root1] += size_[root2];
			potential_delta_[root2] = potential_delta_[index1] - (weight + potential_delta_[index2]);
			parent_[root2] = root1;
		}
		return true;
	}

	// indexが含まれる木の要素数を返す
	int calcSize(const int index)
	{
		return size_[calcRoot(index)];
	}

	// potential(index1) - potential(index2)を返す
	// 根が異なる場合はLLONG_MAXを返す
	int64_t potential(const int index1, const int index2)
	{
		if (calcRoot(index1) != calcRoot(index2))
			return LLONG_MAX;
		return potential_delta_[index1] - potential_delta_[index2];
	}
};

bool solve()
{
	int N, M;
	scanf("%d%d", &N, &M);
	UnionFindWithPotential ufp(N);
	for (int i{}; i < M; i++)
	{
		int L, R;
		long long D;
		scanf("%d%d%lld", &L, &R, &D);
		L--; R--;
		if (!ufp.unite(R, L, D)) return false;
	}
	return true;
}

int main()
{
	if (solve()) puts("Yes");
	else puts("No");

	return 0;
}