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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Scanner;
import java.util.Arrays;

class Main {
    static final Scanner sc = new Scanner(System.in);
        
    static WeightedUnionFind uf;
    static int n, m;
    static int l, r, d;


    public static void main(String[] args) {
        n = sc.nextInt();
        m = sc.nextInt();

        uf = new WeightedUnionFind(n + 1);

        for (int i = 0; i < m; i++) {
            l = sc.nextInt();
            r = sc.nextInt();
            d = sc.nextInt();

            if (uf.isSameGroup(l, r) &&
                uf.diff(l, r) != d) {
                System.out.println("No");
                return;
            }
            else {
                uf.unite(l, r, d);
            }
        }

        System.out.println("Yes");
    }
}


/**
 * 重み付きUnionFind木.
 * アーベル群まで拡張したいけどパフォが気になるので保留
 * ArrayList\<Integer\>がint[]よりだいぶ遅い
 */
class WeightedUnionFind {
    /** 親を示す 自身が根の場合は -(集合の大きさ) を持つ */
    private int[] parent;
    
    /** ノードの重み (親との差分) */
    private int[] diff_weight;

    public WeightedUnionFind(int nmemb) {
        parent = new int[nmemb];
        diff_weight = new int[nmemb];

        Arrays.fill(parent, -1);
        Arrays.fill(diff_weight, 0);
    }

    /** 根を求める */
    private int root(int x) {
        if (parent[x] < 0) {
            return x;
        }

        // 累積和的に重みを再計算
        int r = root(parent[x]);
        diff_weight[x] += diff_weight[parent[x]];
        parent[x] = r;

        return parent[x];
    }

    private int weight(int x) {
        root(x);
        return diff_weight[x];
    }

    public int diff(int x, int y) {
        return weight(y) - weight(x);
    }
    
    public boolean isSameGroup(int x, int y) {
        return root(x) == root(y);
    }

    /**
     * x->yの重みをwとして併合.
     * 即ち, weight(y) = weight(x) + w が成り立つ
     * x, yが既に同じグループのときは何もしない
     */
    public void unite(int x, int y, int w) {
        if (isSameGroup(x, y)) {
            return;
        }

        w += weight(x);
        w -= weight(y);
        x = root(x);
        y = root(y);
    
        parent[y] = x;
        diff_weight[y] = w;
    }

    public int groubSize(int x) {
        return -parent[root(x)];
    }
}