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,unused_imports,dead_code)]
use std::str::FromStr;
use std::collections::{HashMap,HashSet,VecDeque,BinaryHeap};
use std::cmp::{max,min,Reverse,Ordering};

use proconio::input;
use proconio::marker::{Chars,Usize1,Isize1};

fn comp_f64(a: &f64, b: &f64) -> Ordering {
    if a < b {
        return Ordering::Less;
    } else if a > b {
        return Ordering::Greater;
    }
    Ordering::Equal
}

fn main() {
    input!{
        N:usize,
        M:usize,
        LRD:[(Usize1,Usize1,isize);M],
    }
    let mut E = vec![HashMap::new();N];

    for &(l,r,d) in &LRD{
        E[l].insert(r,d);
        E[r].insert(l,-d);
    }

    let INF = 1e15 as isize;
    let mut D = vec![-INF;N];
    for i in 0..N{
        if D[i]==-INF{bfs_dist_weighted_tree(N, &E, i, &mut D);}
    }
    let mut ans = true;
    for &(l,r,d) in &LRD{
        ans &= D[r]-D[l] == d;
    }
    println!("{}",if ans {"Yes"} else {"No"});
}

fn bfs_dist_weighted_tree(_N:usize,E:&Vec<HashMap<usize,isize>>,start:usize,d:&mut Vec<isize>){
        let mut q = VecDeque::new();
        q.push_back(start);
        d[start] = 0;
        while let Some(i) = q.pop_front(){
            for (&j,&c) in E[i].iter(){
                if d[j]!= -1e15 as isize{continue;}
                d[j] = d[i]+c;
                q.push_back(j);
            }
        }
    }