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.Scanner;

class D {
    static class E {
        final int to;
        final int d;

        public E(int to, int d) {
            this.to = to;
            this.d = d;
        }
    }

    static class V extends ArrayList<E> {
    }

    public static void main(String... args) {
        final Scanner sc = new Scanner(System.in);
        final int N = sc.nextInt();
        final int M = sc.nextInt();
        final int[] L = new int[M];
        final int[] R = new int[M];
        final int[] D = new int[M];
        for (int i = 0; i < M; i++) {
            L[i] = sc.nextInt() - 1;
            R[i] = sc.nextInt() - 1;
            D[i] = sc.nextInt();
        }
        final V[] vs = new V[N];
        for (int i = 0; i < N; i++)
            vs[i] = new V();
        for (int i = 0; i < M; i++) {
            vs[L[i]].add(new E(R[i], D[i]));
            vs[R[i]].add(new E(L[i], -D[i]));
        }
        final boolean[] visited = new boolean[N];
        final int[] ds = new int[N];
        boolean success = true;
        for (int i = 0; i < N && success; i++)
            if (!visited[i]) {
                visited[i] = true;
                success &= check(i, vs, visited, ds);
            }
        System.out.println(success ? "Yes" : "No");
    }

    static boolean check(int idx, V[] vs, boolean[] visited, int[] ds) {
        for (final E e : vs[idx])
            if (visited[e.to]) {
                if (ds[idx] + e.d != ds[e.to])
                    return false;
            } else {
                visited[e.to] = true;
                ds[e.to] = ds[idx] + e.d;
                if (!check(e.to, vs, visited, ds))
                    return false;
            }
        return true;
    }
}
public class Main {
    public static void main(String... args) {
        D.main(args);
    }
}