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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <algorithm>
#include <sstream>
#include <vector>

struct elm
{
    int L;
    int R;
    int D;
};

const int EMPTY=1000000001;
int main(void)
{
    int N, M;
    std::cin >> N >> M;
    std::vector<int> h(N+1, EMPTY);
    std::vector<elm> m(M), mr;
    for(int i=0; i<M; ++i)
        std::cin >> m[i].L >> m[i].R >> m[i].D;
    std::sort(m.begin(), m.end(), [](elm &l, elm &r){return l.L < l.R;});

    if(m.empty())
    {
        std::cout << "Yes" << std::endl;
        return 0;
    }
    h[m.back().L] = 0;
    h[m.back().R] = m.back().D;
    m.pop_back();
    size_t m_ps = 0;
    while(!m.empty())
    {
        if(m.size() == m_ps)
            break;
        m_ps = m.size();
        for(auto &e : m)
        {
            if(h[e.L] != EMPTY)
            {
                if(h[e.R] != EMPTY)
                {
                    if(h[e.R] - h[e.L] != e.D)
                    {
                        std::cout << "No" << std::endl;
                        return 0;
                    }
                }
                else
                {
                    h[e.R] = h[e.L] + e.D;
                }
            }
            else
            {
                if(h[e.R] != EMPTY)
                {
                    h[e.L] = h[e.R] - e.D;
                }
                else
                {
                    mr.push_back(e);
                }
            }
        }
        std::swap(m, mr);
        mr.clear();
    }
    std::cout << "Yes" << std::endl;
    return 0;
}