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)

pub fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

pub fn read_vec2<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

pub fn read_col<T: std::str::FromStr>(n: usize) -> Vec<T> {
    (0..n).map(|_| read()).collect()
}

struct UnionFind {
    parent: Option<usize>,
    dist: u32,
}

impl Clone for UnionFind {
    fn clone(&self) -> Self {
        UnionFind {
            parent: self.parent.clone(),
            dist: self.dist,
        }
    }
}

fn union(l: usize, r: usize, d: u32, uf: &mut Vec<UnionFind>) -> bool {
    let (lr, ld) = root(l, uf);
    let (rr, rd) = root(r, uf);

    if lr == rr {
        if ld + d == rd {
            return true;
        } else {
            return false;
        }
    }
    if ld + d >= rd {
        uf[rr].parent = Some(lr);
        uf[rr].dist = ld + d - rd;
    } else {
        uf[lr].parent = Some(rr);
        uf[lr].dist = rd - ld - d;
    }
    true
}

fn root(a: usize, uf: &mut Vec<UnionFind>) -> (usize, u32) {
    if let Some(b) = uf[a].parent {
        let (r, d) = root(b, uf);
        uf[a].dist += d;
        uf[a].parent = Some(r);
        (r, uf[a].dist)
    } else {
        (a, 0)
    }
}

fn main() {
    let nm: Vec<usize> = read_vec();
    let n = nm[0];
    let m = nm[1];

    let mut uf: Vec<UnionFind> = vec![
        UnionFind {
            parent: None,
            dist: 0,
        };
        n
    ];

    for _ in 0..m {
        let lrd: Vec<usize> = read_vec();
        let l = lrd[0] - 1;
        let r = lrd[1] - 1;
        let d = lrd[2] as u32;
        if !union(l, r, d, &mut uf) {
            println!("No");
            return;
        }
    }

    println!("Yes");
}