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: Wrong Answer)

use std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn dfs(G: &Vec<Vec<(usize, u64)>>, memo: &mut Vec<Option<u64>>, x: usize, pos: u64) -> bool {
    if let Some(p) = memo[x] {
        if p != pos {
            return false;
        }
    }
    memo[x] = Some(pos);
    for &(y, d) in G[x].iter() {
        if !dfs(G, memo, y, pos + d) {
            return false;
        }
    }
    true
}

fn main() {
    let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = line[0];
    let m = line[1];

    let mut G = vec![Vec::new(); n];
    let mut root = vec![true; n];

    for _ in 0..m {
        let line: Vec<u64> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
        let l = (line[0] - 1) as usize;
        let r = (line[1] - 1) as usize;
        let d = line[2];
        G[l].push((r, d));
        root[r] = false;
    }

    let mut memo = vec![None; n];
    let mut ans = true;
    for i in 0..n {
        if root[i] {
            ans &= dfs(&G, &mut memo, i, 0);
        }
    }

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