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::io::{stdin, Read, StdinLock};
use std::str::FromStr;

#[allow(dead_code)]
struct Scanner<'a> {
    cin: StdinLock<'a>,
}
#[allow(dead_code)]
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }

    fn read<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }

    fn input<T: FromStr>(&mut self) -> T {
        self.read().unwrap()
    }
}
#[allow(dead_code)]
type Graph = Vec<Vec<(usize, i64)>>;
fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);

    let (n, m): (i64, i64) = (sc.input(), sc.input());


    let mut graph: Graph = vec![vec![]; n as usize];
    for _ in 0..m {
        let (mut l, mut r, d): (i64, i64, i64) = (sc.input() , sc.input(), sc.input());
        l -= 1; r -=1;
        graph[l as usize].push((r as usize, d));
        graph[r as usize].push((l as usize, -d));
    }
    println!("{}", if dfs(&graph) { "Yes"} else { "No" });
}

fn dfs(graph: &Graph) -> bool {
    let mut dist :Vec<Option<i64>> = vec![None; graph.len()];
    let mut stack = vec![];

    for i in 0..graph.len() {
        if dist[i] != None { continue; }

        dist[i] = Some(0);
        stack.push((i as usize, 0 as usize));
        while let Some((cur, index)) = stack.pop() {
            if index == graph[cur].len() { continue; }

            stack.push((cur, index + 1));
            let (next, d) = graph[cur][index];

            if dist[next] != None {
                if dist[cur].unwrap() as i64 + d
                    != dist[next].unwrap() as i64 { return false; }
            } else {
                dist[next] = Some(dist[cur].unwrap() as i64 + d as i64);
                stack.push((next, 0));
            }
        }
    }
    true
}