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: Accepted)

#include <bits/stdc++.h>
#define all(x) begin(x), end(x)
#define _ << ' ' <<
using namespace std;
using ll = long long;

int dest[100005];
vector<pair<int, int>> adj[100005];
int vis[100005];
vector<int> sol;

void dfs(int x)
{
    if (vis[x])
        return;

    vis[x] = 1;
    for (auto p : adj[x])
        dfs(p.first);
    sol.push_back(x);
}

int main()
{
    int n, m;
    cin >> n >> m;
    vector<tuple<int, int, int>> v;
    for (int i = 0; i < m; ++i)
    {
        int l, r, d;
        cin >> l >> r >> d;
        adj[l].emplace_back(r, d);
    }
    for (int i = 1; i <= n; ++i)
        dfs(i);

    reverse(all(sol));
    for (int x : sol)
    {
        for (auto p : adj[x])
        {
            int y, d;
            tie(y, d) = p;
            if (!dest[x]) dest[x] = 1;
            if (!dest[y]) dest[y] = dest[x] + d;
            if (dest[y] != dest[x] + d)
                return cout << "No", 0;
        }
    }
    cout << "Yes";
}