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>
#include<math.h>
#include<string.h>
//1 2 3 4 100 1
/*
配列を持っているa
四こに分ける
どれぞれの総和をおんなじくらいへ
配列の総和を半分にカットする関数

1回めのカット
右に
*/
#define N_INFTY -2147483648



int main(void){
    int N,M;
    scanf("%d %d",&N,&M);
    int L[M],R[M],D[M],p_p[N],man[N];
    int i;
    for(i=0;i<N;i++){
        man[i] = 0;
    }
    for(i=0;i<M;i++){
        scanf("%d %d %d",&L[i],&R[i],&D[i]);
        if(man[L[i]]==0 && man[R[i]] == 0){
            man[L[i]] = 1;
            man[R[i]] = 1;
            p_p[L[i]] = 0;
            p_p[R[i]] = D[i];
        }else if(man[L[i]]==0 && man[R[i]] == 1){
            man[L[i]] = 1;
            p_p[L[i]] = p_p[R[i]] - D[i];
        }else if(man[L[i]]==1 && man[R[i]] == 0){
            man[R[i]] = 1;
            p_p[R[i]] = p_p[L[i]] + D[i];
        }else if(man[L[i]]==1 && man[R[i]] == 1){
            if(p_p[R[i]]-p_p[L[i]] != D[i]){
                printf("No\n");
                break;
            }
        }
        if(i==M-1){
            printf("Yes\n");
        }
    }

    return 0;
}