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

//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)

//printf
#define PRINTD(d) printf("%d\n",(d))
#define PRINTS(s) printf("%s\n",(s))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

typedef long long ll;

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int a, int b) {
	return (a)>(b) ? (a) : (b);
}

int x[100001];

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	REPS(i, n)x[i] = -99999999;
	REPS(i, m) {
		int l, r, d;
		scanf("%d %d %d", &l, &r, &d);
		if (x[l] == -99999999 && x[r] == -99999999) {
			x[l] = 0;
			x[r] = d;
		}
		else {
			if (x[l] == -99999999) {
				x[l] = x[r] - d;
			}
			else if(x[r]== -99999999){
				x[r] = x[l] + d;
			}
			else {
				if (x[r] - x[l] != d) {
					PRINTS("No");
					return 0;
				}
			}
		}
	}
	int min = 0;
	REPS(i, n) {
		if (x[i]>-99999999 && min > x[i])min = x[i];
	}
	REPS(i, n) {
		x[i] -= min;
		if (x[i] > 1000000000) {
			PRINTS("No");
			return 0;
		}
	}
	PRINTS("Yes");
}