Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in JAVA):

Solution:

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

public class Main {
    int n, m;
    int[] ls, rs, ds;
    long[] xs;
    ArrayList<ArrayList<Edge>> es;
    boolean flg;
    public static void main(String[] args) {
        Main m = new Main();
        m.read();
        m.solve();
    }

    private void read() {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        ls = new int[m];
        rs = new int[m];
        ds = new int[m];
        es = new ArrayList<>(n);
        for (int i = 0; i < n; i++)
            es.add(new ArrayList<>());
        for (int i = 0; i < m; i++) {
            ls[i] = sc.nextInt();
            rs[i] = sc.nextInt();
            ds[i] = sc.nextInt();
            es.get(ls[i]- 1).add(new Edge(rs[i]- 1, ds[i]));
            es.get(rs[i]-1).add(new Edge(ls[i]-1, -ds[i]));
        }
    }

    private void solve() {
        xs = new long[n];
        Arrays.fill(xs, Long.MAX_VALUE);
        flg = true;
        for (int i = 0; i < n; i++) {
            if (xs[i] == Long.MAX_VALUE && flg) {
                xs[i] = 0;
                dfs(i);
            }
        }
        if (flg)
            System.out.println("Yes");
        else
            System.out.println("No");
    }

    private void dfs(int cur) {
        ArrayList<Edge> c_es = es.get(cur);
        for (Edge e: c_es) {
            if (xs[e.to] == Long.MAX_VALUE) {
                xs[e.to] = e.cost + xs[cur];
                dfs(e.to);
            }
            else {
                long check = e.cost + xs[cur];
                flg &= xs[e.to] == check;
            }
        }
    }

    private class Edge {
        int to, cost;
        Edge(int t, int c) {
            to = t;
            cost = c;
        }
    }
}