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)

#![allow(non_snake_case,unused_imports)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;


fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let N:usize = iter.next().unwrap().parse().unwrap();
    let M:usize = iter.next().unwrap().parse().unwrap();
    let mut E = (0..N).map(|_| HashMap::new()).collect::<Vec<_>>();

    for _ in 0..M{
        let l = iter.next().unwrap().parse::<usize>().unwrap() -1;
        let r = iter.next().unwrap().parse::<usize>().unwrap() -1;
        let d = iter.next().unwrap().parse::<isize>().unwrap();
        E[l].insert(r,d);
        E[r].insert(l,-d);
    }

    let mut ans = true;

    let INF = 1_000_100_000;
    let mut D = (0..N).map(|_| INF).collect::<Vec<_>>();
    for i in 0..N{
        if D[i] != INF {continue;}
        D[i] = 0;
        let mut q = vec![i];
        while q.len()>0{
            let j = q.pop().unwrap();
            let dj = D[j];
            for (&k,&d) in &E[j]{
                if D[k] == INF{
                    D[k] = dj + d;
                    q.push(k);
                } else {
                    ans = ans && ((dj + d) == D[k]);
                }
            }
        }
    }
    println!("{}",if ans {"Yes"} else {"No"});
}