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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int dis[100010], par[100010];
int flag = 1;

int find(int x){
    if(par[x] == x){
        return x;
    }else{
        int r = find(par[x]);
        dis[x] = dis[par[x]] + dis[x];
        par[x] = r;
        return par[x];
    }
}

void unite(int x, int y, int z){
    int fx = find(x);
    int fy = find(y);

    if(fx == fy){
        if(abs(dis[x] - dis[y]) != z){
            flag = 0;
        }
    }else{
        par[fy] = fx;
        dis[fy] = dis[x] + z - dis[y];
    }
}

int main(void){

    int N, M;

    scanf("%d%d", &N, &M);

    for(int i = 0; i < N; i++){
        par[i] = i;
    }
    
    for(int i = 0; i < M; i++){
        int l, m, d;
        scanf("%d%d%d", &l, &m, &d);
        unite(l, m, d);
    }

    if(flag == 1){
        printf("Yes\n");
    }else{
        printf("No\n");
    }

    return 0;
}