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::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 uf = UnionFind::new(n);
    let mut ok = true;
    for (l, r, d) in lrd {
        let l = l - 1;
        let r = r - 1;
        if uf.is_same(l, r) {
            if uf.diff(l, r) != d {
                ok = false;
                break;
            }
        } else {
            uf.merge(l, r, d);
        }
    }

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

struct UnionFind {
    parent: Vec<i64>,
    weight: Vec<i64>,
}
impl UnionFind {
    fn new(n: usize) -> Self {
        UnionFind {
            parent: vec![-1; n],
            weight: vec![0; n],
        }
    }

    fn root(&mut self, i: usize) -> usize {
        if self.parent[i] < 0 {
            i
        } else {
            let p = self.parent[i] as usize;
            let p = self.root(p);
            self.parent[i] = p as i64;
            self.weight[i] += self.weight[self.parent[i] as usize];
            p
        }
    }

    fn weight(&mut self, i: usize) -> i64 {
        self.root(i);
        self.weight[i]
    }

    fn is_same(&mut self, i: usize, j: usize) -> bool {
        self.root(i) == self.root(j)
    }

    fn merge(&mut self, i: usize, j: usize, w: i64) -> bool {
        let w = w + self.weight(i) - self.weight(j);
        let i = self.root(i);
        let j = self.root(j);
        if i == j {
            false
        } else {
            self.parent[j] = i as i64;
            self.weight[j] = w;
            true
        }
    }

    fn diff(&mut self, i: usize, j: usize) -> i64 {
        self.weight(j) - self.weight(i)
    }
}