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: Runtime Error)

use std::vec::Vec;
use std::cmp::*;
use std::collections::HashSet;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let mut hash: HashSet<usize> = HashSet::new();
  let (n, m) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };
  let mut v: Vec<(usize, usize, i64)> = Vec::new();
  for _ in 0..m {
    let t: Vec<usize> = read_vec();
     v.push((t[0] - 1, t[1] - 1, t[2] as i64));
     hash.insert(t[0] - 1);
     hash.insert(t[1] - 1);
  }

  let mut posit: Vec<i64> = vec![std::i64::MAX; n];

  let g = Graph::new(n, &v);
  for i in hash {
    if posit[i] == std::i64::MAX {
      posit[i] = 0;
      g.dfs(i, &mut posit);
    }
  }

  let mut b = true;
  for &e in &v {
    if posit[e.1] - posit[e.0] != e.2 {
      b = false;
      break;
    }
  }

  if b {
    println!("Yes");
  } else {
    println!("No");
  }
}


#[derive(Debug, Clone)]
struct Graph {
  n: usize,                            // number of nodes
  graph: Vec<Vec<usize>>,              // first: from node, second: to node, third: cost
  cost: Vec<Vec<i64>>,                   // cost matrix
}

impl Graph {
  // constructor
  fn new(n: usize, v: &Vec<(usize, usize, i64)>) -> Graph {
    let mut g: Vec<Vec<usize>>  = vec![Vec::new(); n];
    let mut c: Vec<Vec<i64>> = vec![vec![0; n]; n];
    for &(from, to, cost) in v {
      g[from].push(to);
      c[from][to] = cost;
      g[to].push(from);
      c[to][from] = - cost;
    }

    Graph {
        n: n,
        graph: g,
        cost: c,
    }
  }

  // depth first search
  fn dfs(&self, start: usize, v: &mut Vec<i64>) {
    let mut visit: Vec<bool> = vec![false; self.n];    // record already visit to node
    let mut stack: Vec<usize> = Vec::new();

    // init
    visit[start] = true;
    stack.push(start);

    while !stack.is_empty() {
      let node: usize = match stack.pop() {
        None => panic!("unifined"),
        Some(i) => i,
      };


      for &e in &self.graph[node] {
        if !visit[e] {
          stack.push(e);
          visit[node] = true;  // already visit node v, so mark
          v[e] = self.cost[node][e] + v[node];
        }
      }
    }

  }
}