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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#define MAX 100000000

typedef struct{
    int L;
    int R;
    int D;
}info;

int comp(const void *a, const void *b){
    info s_a = *(info*)a, s_b = *(info*)b;
    if(s_a.L == s_b.L){
        return s_a.R - s_b.R;
    }else{
        return s_a.L - s_b.L;
    }
}

int main(){
    int N, M;
    int i;
    scanf("%d %d", &N, &M);
    info data[M];
    int position[N];

    for(i = 0; i < M; ++i){
        scanf("%d %d %d", &data[i].L, &data[i].R, &data[i].D);
    }
    
    if(M > 0){
        qsort(data, M, sizeof(info), comp);
        position[data[0].L - 1] = 0;

        for(i = 0; i < N; ++i){
            if(i != data[0].L - 1){
                position[i] = MAX;
            }
        }
    }
    /*
    for(i = 0; i < M; ++i){
        printf("(L, R, D) = (%d %d %d)\n", data[i].L, data[i].R, data[i].D);
    }
    */

    char flag = 1;
    int num;

    for(num = 0; num < 2; num++){
        for(i = 0; i < M; ++i){
            int l = position[data[i].L - 1], r = position[data[i].R - 1];
            if(l == MAX && r != MAX){
                position[data[i].L - 1] = r - data[i].D;
            }else if(r == MAX && l != MAX){
                position[data[i].R - 1] = l + data[i].D;
            }else if(r != MAX && l != MAX){
                flag &= (r - l == data[i].D);
                if(!flag){
                    break;
                }
            }
        }
    }

    if(flag){
        printf("Yes\n");
    }else{
        printf("No\n");
    }
    /*
    for(i = 0; i < N; ++i){
        if(position[i] != INT32_MAX){
            printf("%d : %d\n", i, position[i]);
        }
    }
    */

    return 0;
}