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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
// 重み付きUnion-Find：同じグループに属するかどうかの判定と根からの重みを保持する
class UnionFind2
{
    private $par; // par[$i]:$iの親 自分が根の場合は自身のindexとなる
    private $size; // size[$i]:$iが属するグループの頂点の数
    private $rank; // rank[$i]:$iの重み
    private $diff_weight; // diff_weight[$i]:$iの重みの差

    function __construct($n, $sum_unity = 0)
    {
        for ($i = 0; $i < $n; $i++) {
            $this->par[$i] = $i; // 最初は全てが根
            $this->size[$i] = 1; // 最初は全て1
            $this->rank[$i] = 0; // 最初は全て0
            $this->diff_weight[$i] = $sum_unity;
        }
    }

    function root($x)
    { // 根を返す
        if ($this->par[$x] === $x) return $x; // 根の場合
        $r = $this->root($this->par[$x]);
        $this->diff_weight[$x] += $this->diff_weight[$this->par[$x]]; // 全ての枝の重みの差を根との重みの差にする
        return $this->par[$x] = $r; // 全ての枝の親を根にする
    }

    function weight($x)
    {
        $this->root($x);
        return $this->diff_weight[$x];
    }

    function unite($x, $y, $w)
    { // $xと$yの木を併合
        $w += $this->weight($x);
        $w -= $this->weight($y);
        $ix = $this->root($x);
        $iy = $this->root($y);
        if ($ix === $iy) return;
        if ($this->rank[$ix] < $this->rank[$iy]) {
            list($ix, $iy) = [$iy, $ix];
            $w = -$w;
        }
        if ($this->rank[$ix] === $this->rank[$iy]) $this->rank[$ix]++;
        $this->size[$ix] += $this->size[$iy];
        $this->par[$iy] = $ix; // $yの根を$xの根に付ける
        $this->diff_weight[$iy] = $w;
    }

    function isSame($x, $y)
    { // $xと$yの根が同じか
        return $this->root($x) === $this->root($y);
    }

    function diff($x, $y)
    {
        return $this->weight($y) - $this->weight($x);
    }

    function size($x)
    {
        return $this->size[$this->root($x)];
    }
}

list($n, $m) = ints();
$tree = new UnionFind2($n);
for ($i = 0; $i < $m; $i++) {
    list($l, $r, $d) = ints();
    $l--;
    $r--;
    if ($tree->isSame($l, $r)) {
        if ($tree->diff($l, $r) !== $d) exit('No');
    } else {
        $tree->unite($l, $r, $d);
    }
}
echo 'Yes';

function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}