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)

#![allow(non_snake_case)]

use proconio::input;
use proconio::marker::Usize1;
use std::vec::Vec;

fn main() {
  input! {
    n: usize,
    m: usize,
    lrd: [(Usize1, Usize1, isize); 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 nanachi = vec![None; n];
  let mut visited = vec![false; n];
  for i in 0..n {
    if !dfs(i, &g, &mut nanachi, &mut visited) {
      println!("No");
      return;
    }
  }
  println!("Yes");
}

fn dfs(
  v: usize,
  g: &Vec<Vec<(usize, isize)>>,
  nanachi: &mut Vec<Option<isize>>,
  visited: &mut Vec<bool>,
) -> bool {
  if visited[v] {
    return true;
  }
  if !nanachi[v].is_some() {
    nanachi[v] = Some(0);
  }
  let x = nanachi[v].unwrap();
  for &(w, d) in &g[v] {
    if let Some(y) = nanachi[w] {
      if x + d != y {
        return false;
      }
    } else {
      nanachi[w] = Some(x + d);
      if !dfs(w, g, nanachi, visited) {
        return false;
      }
    }
  }
  true
}