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(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      n: usize,
      m: usize,
      lrd: [(Usize1, Usize1, i64); m]
   }

   let mut to = vec![vec![]; n];
   for &(l, r, d) in lrd.iter() {
      to[l].push((r, d));
      to[r].push((l, -d));
   }

   let mut d = BTreeMap::<usize, i64>::new();
   for i in 0..n {
      if d.contains_key(&i) {
         continue;
      }
      if !dfs(i, 0i64, &mut d, &to) {
         println!("No");
         return;
      }
   }
   println!("Yes");
}

fn dfs(v: usize, now: i64, d: &mut BTreeMap<usize, i64>, to: &Vec<Vec<(usize, i64)>>) -> bool {
   d.insert(v, now);
   let mut b = true;
   for &(u, x) in to[v].iter() {
      if let Some(&v) = d.get(&u) {
         if v != now + x {
            b = false;
         }
      } else {
         b = dfs(u, now + x, d, to);
      }

      if !b {
         break;
      }
   }
   b
}