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)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let m: usize = input();
    let lrd: Vec<(usize, usize, i64)> = (0..m)
        .map(|_| {
            let l = input();
            let r = input();
            let d = input();
            (l, r, d)
        })
        .collect();

    let mut g = vec![vec![]; n + 1];
    for &(l, r, d) in &lrd {
        g[l].push((r, d));
        g[r].push((l, -d));
    }

    let mut v = vec![false; n + 1];
    let mut x = vec![0; n + 1];
    for i in 1..n + 1 {
        if !v[i] {
            v[i] = true;
            dfs(i, &mut x, &mut v, &g);
        }
    }

    let mut ok = true;
    for (l, r, d) in lrd {
        if x[r] - x[l] != d {
            ok = false;
        }
    }

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

fn dfs(i: usize, x: &mut Vec<i64>, v: &mut Vec<bool>, g: &Vec<Vec<(usize, i64)>>) {
    for &(j, d) in &g[i] {
        if !v[j] {
            v[j] = true;
            x[j] = x[i] + d;
            dfs(j, x, v, g);
        }
    }
}