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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>

#define ARC(i,a,b) for(int i=(a);i<(b);++i)
#define ARCH(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define RC(i,n) ARC(i,0,n)
#define RCH(i,n) ARCH(i,n,0)
#define SRC(i,s) for(Int i=0;(s)[i];++i)
#define ATH(a,b) ((a)>(b)?(a):(b))
#define OTH(a,b) ((a)<(b)?(a):(b))
#define ANTH(a,b) {if((a)<(b))(a)=(b);}
#define ONTH(a,b) {if((a)>(b))(a)=(b);}
#define ACH printf
#define ECH scanf
#define AC putchar
#define EC getchar
#define _ AC('\n');

int main() {
	int n, m;
	static int tr[2][110000];
	int ans;
	ECH("%d%d", &n, &m);
	ans = n;
	RC(i, n) tr[1][i] = i;
	RC(i, n) tr[2][i] = 0;
	RC(i, m){
		int l, r, d;
		ECH("%d%d%d", &l, &r, &d);
		--l;
		--r;
		while(tr[1][l] != tr[1][tr[1][l]]){
			tr[2][l] += tr[2][tr[1][l]];
			tr[1][l] = tr[1][tr[1][l]];
		}
		while(tr[1][r] != tr[1][tr[1][r]]){
			tr[2][r] += tr[2][tr[1][r]];
			tr[1][r] = tr[1][tr[1][r]];
		}
		if(tr[1][tr[1][l]] == tr[1][tr[1][r]]){
			if(tr[2][r] - tr[2][l] != d) --ans;
		}else{
			tr[2][tr[1][r]] = tr[2][l] + d - tr[2][r];
			tr[1][tr[1][r]] = tr[1][tr[1][l]];
		}
		RC(i, n) ACH("%d %d %d\n", i, tr[1][i], tr[2][i]);
	}
	ACH(ans == n ? "Yes\n" : "No\n");
}