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 <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <queue>
#include <utility>
#include <tuple>
#include <climits>
using namespace std;
using Int = long long;

int N, M;
struct info { int L, R, D; };
vector<info> dat;
vector<int> pos;
vector<bool> walked;

bool walk(int n, int i)
{
    walked[n] = true;
    int j = i;
    while (j < dat.size() && dat[j].L == dat[i].L) j++;
    for (int k = i; k < j; k++) {
        int m = dat[k].R;
        if (pos[m] == INT_MIN) {
            pos[m] = pos[n] + dat[k].D;
        } else if (pos[n] + dat[k].D != pos[m]) {
            return true;
        }
    }
    for (int k = i; k < j; k++) {
        int m = dat[k].R;
        if (!walked[m]) {
            if (walk(m, partition_point(dat.begin(), dat.end(), [=](info &a){return a.L<m;}) - dat.begin()))
                return true;
        }
    }
    return false;
}

int main()
{
    cin >> N >> M;
    pos.resize(N+1);
    walked.resize(N+1);
    fill(pos.begin(), pos.end(), INT_MIN);
    for (int i = 0; i < M; i++) {
        int l, r, d;
        cin >> l >> r >> d;
        dat.push_back({l, r, d});
        dat.push_back({r, l, -d});
    }
    sort(dat.begin(), dat.end(), [](info &a, info &b){return a.L < b.L;});
    for (int i = 0; i < dat.size(); i++) {
        int L = dat[i].L;
        if (pos[L] != INT_MIN) continue;
        pos[L] = 0;
        if (walk(L, i)) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
    return 0;
}