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: Time Limit Exceeded)

use std::io::{self, BufRead};
fn main() {
    let stdin = io::stdin();
    let mut handle = stdin.lock();
    let mut buf = String::new();
    handle.read_line(&mut buf).unwrap();
    let tmp = buf.trim().split_whitespace().map(|x| x.parse::<usize>().unwrap()).collect::<Vec<_>>();
    let N = tmp[0];
    let M = tmp[1];
    buf.clear();

    let LRD = {
        let mut ret = Vec::with_capacity(M);
        for i in 0..M {
            handle.read_line(&mut buf).unwrap();
            let tmp = buf.trim().split_whitespace().map(|x| x.parse::<usize>().unwrap()).collect::<Vec<_>>();
            ret.push((tmp[0] - 1, tmp[1] - 1, tmp[2] as i64));
            buf.clear();
        }
        ret
    };

    let mut edges = Vec::new();
    for _ in 0..N {
        edges.push(vec![]);
    }
    for &(l, r, d) in &LRD {
        edges[l].push((r, d));
        edges[r].push((l, -d));
    }

    use std::collections::{HashSet, VecDeque};
    let mut visited = HashSet::new();
    for i in 0..N {
        if visited.contains(&i) {
            continue;
        }
        let mut rel_pos = vec![None; N];
        let mut queue = VecDeque::new();
        queue.push_back((i, 0));
        while let Some((n, pos)) = queue.pop_front() {
            if visited.contains(&n) {
                continue;
            }
            rel_pos[n] = Some(pos);
            visited.insert(n);
            for &(to, d) in &edges[n] {
                if rel_pos[to].is_some() && pos + d != rel_pos[to].unwrap() {
                    println!("No");
                    return;
                }
                queue.push_back((to, pos + d));
            }
        }
    }
    println!("Yes");
}