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: Time Limit Exceeded)

#![allow(unused_imports, unused_variables, dead_code)]
use std::io::{stdin, Read, StdinLock};
use std::str::FromStr;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

// https://beta.atcoder.jp/contests/arc065/tasks/arc065_b

fn bfs(
    cmap: &Vec<Vec<i64>>
) -> bool {
    let mut q: VecDeque<usize> = VecDeque::new();
    let mut visited: [bool; 100005] = [false; 100005];
    let mut node: [i64; 100005] = [0; 100005];
    for n in 0..cmap.len() {
        if visited[n] { continue; }
        visited[n] = true;
        node[n] = 0;
        q.push_back(n);
        while !q.is_empty() {
            let &c = q.front().unwrap();
            for i in 0..cmap.len() {
                if i == c { continue; }
                let c_to_i = cmap[c][i];
                if c_to_i == 110000 { continue; }
                if visited[i] { // 既に決めた場所なのでチェック
                    if node[c] + c_to_i == node[i] { continue; } //OK
                    else { return false; }
                }
                // きたことないのでNodeの値を決める
                node[i] = node[c] + c_to_i;
                visited[i] = true;
                q.push_back(i);
            }
            q.pop_front();
        }
    }
    return true;
}

fn main() {
    let cin = stdin(); let cin = cin.lock(); let mut sc = Scanner::new(cin);

    let (n, m): (usize, usize) = (sc.read(), sc.read());
    // NxNの２次元配列
    let mut lrd: Vec<Vec<i64>> = vec![vec![110000; n]; n];

    for i in 0..m {
        let (l, r, d): (usize, usize, i64) = (sc.read(), sc.read(), sc.read());
        let l = l - 1; let r = r - 1;
        lrd[l][r] = d;
        lrd[r][l] = -d;
    }

    if !bfs(&lrd) { println!("No"); }
    else { println!("Yes"); }
}

struct Scanner<'a> {
    cin: StdinLock<'a>,
}
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }
 
    fn read1<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }
 
    fn read<T: FromStr>(&mut self) -> T {
        self.read1().unwrap()
    }
}