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 (1.15.1) (judge verdict: Accepted)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let m: usize = itr.next().unwrap().parse().unwrap();

    let mut g = vec![Vec::new(); n];
    for _ in 0..m {
        let l = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        let r = itr.next().unwrap().parse::<usize>().unwrap() - 1;
        let c: i64 = itr.next().unwrap().parse().unwrap();
        // l, rの最短距離がc
        // マイナスの辺をはって各頂点ごとに見る
        // 辺の数は
        g[l].push((r, c));
        g[r].push((l, -c));
    }

    const INF: i64 = 1 << 30;
    let mut d = vec![INF; n];
    let mut q = std::collections::VecDeque::new();

    for i in 0..n {
        if d[i] == INF {
            d[i] = 100000;
        }
        q.clear();
        q.push_back(i);
        while let Some(v) = q.pop_front() {
            for &(nv, c) in g[v].iter() {
                if d[nv] == INF {
                    d[nv] = d[v] + c;
                    q.push_back(nv);
                } else {
                    if d[nv] != d[v] + c {
                        println!("No");
                        return;
                    }
                }
            }
        }
    }

    println!("Yes");
}