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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#define inf 10e9 + 5
typedef long long ll;

int main(void) {
  int n, m, i, j, k;
  scanf("%d%d", &n, &m);
  int l, r, d;
  ll **wei;
  wei = (ll**)malloc(sizeof(ll*) * (n + 1));
  for(i = 0; i <= n; ++i) wei[i] = (ll*)malloc(sizeof(ll) * (n + 1));
  char f1[n + 1];
  for(i = 0; i <= n; ++i) for(j = 0; j <= n; ++j) wei[i][j] = inf;
  for(i = 0; i < m; ++i) {
    scanf("%d%d%d", &l, &r, &d);
    wei[l][r] = d;
  }
  for(i = 1; i <= n; ++i) {
    for(j = 1; j <= n; ++j) {
      for(k = 1; k <= n; ++k) {
        if(wei[i][k] != inf) {
          if(wei[i][j] != inf && wei[j][k] != inf) {
            if(wei[i][j] + wei[j][k] != wei[i][k]) { printf("No"); return 0;}
          }
        } else {
          if(wei[i][j] != inf && wei[j][k] != inf) wei[i][k] = wei[i][j] + wei[j][k];
        }
      }
    }
  }
  printf("Yes");
  for(i = 0; i <= n; ++i) free(wei[i]);
  free(wei);
  return 0;
}