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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::{input, marker::Usize1};

fn main() {
    input!(n: usize, m: usize, lrd: [(Usize1, Usize1, i64); m]);
    let mut g = vec![vec![]; n];
    for &(l, r, d) in &lrd {
        g[l].push((r, d));
        g[r].push((l, -d));
    }
    let mut pos = vec![None; n];
    for i in 0..n {
        if pos[i].is_some() {
            continue;
        }
        pos[i] = Some(0);
        let mut queue = std::collections::VecDeque::from(vec![i]);
        while let Some(x) = queue.pop_front() {
            let px = pos[x].unwrap();
            for &(y, d) in &g[x] {
                if let Some(py) = pos[y] {
                    if px + d != py {
                        println!("No");
                        std::process::exit(0);
                    }
                } else {
                    pos[y] = Some(px + d);
                    queue.push_back(y);
                }
            }
        }
    }
    println!("Yes");
}