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)

#[allow(dead_code)]
struct Utree {
    n: usize,
    par: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
    diff_weight: Vec<i64>,
}
#[allow(dead_code)]
impl Utree {
    fn new(n: usize) -> Utree {
        Utree {
            n: n,
            par: (0..n).collect(),
            rank: vec![0; n],
            size: vec![1; n],
            diff_weight: vec![0; n],
        }
    }
    fn find(&mut self, x: usize) -> usize {
        if self.par[x] == x {
            x
        } else {
            let parent = { self.par[x] };
            let ret = self.find(parent);
            self.diff_weight[x] += self.diff_weight[parent];
            self.par[x] = ret;
            ret
        }
    }
    fn get_weight(&mut self, x: usize) -> i64 {
        self.find(x);
        self.diff_weight[x]
    }
    fn get_size(&mut self, x: usize) -> usize {
        let x_root = self.find(x);
        self.size[x_root]
    }
    fn is_same(&mut self, x: usize, y: usize) -> bool {
        let x_root = self.find(x);
        let y_root = self.find(y);
        x_root == y_root
    }
    fn unite(&mut self, x: usize, y: usize, w: i64) {
        let mut dw = w;
        dw += self.get_weight(x);
        dw -= self.get_weight(y);
        let x_root = self.find(x);
        let y_root = self.find(y);
        if x_root == y_root {
            return;
        };
        if self.rank[x_root] < self.rank[y_root] {
            self.par[x_root] = y_root;
            self.size[y_root] += self.size[x_root];
            self.diff_weight[x_root] -= dw;
        } else {
            self.par[y_root] = x_root;
            self.size[x_root] += self.size[y_root];
            self.diff_weight[y_root] += dw;
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }
}

fn main() {
    use std::io::Read;
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    
    let n: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let m: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let mut utree = Utree::new(n);
    let mut ret = true;
    for _ in 0..m {
        let l: usize = iter.next().unwrap().parse::<usize>().unwrap()-1;
        let r: usize = iter.next().unwrap().parse::<usize>().unwrap()-1;
        let d: i64 = iter.next().unwrap().parse::<i64>().unwrap();
        if utree.is_same(l, r) {
            if utree.get_weight(r) - utree.get_weight(l) != d {
                ret = false;
                break;
            }
        } else {
            utree.unite(l, r, d);
        }
    }

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