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: Accepted)

#include <float.h>
#include <inttypes.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <unistd.h>
#include <math.h>
#ifdef __cplusplus
#include <bits/stdc++.h>
#endif

#define getchar getchar_unlocked
#define putchar putchar_unlocked
int32_t nextint(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); bool s=false; if(c=='-'){s=true;c=getchar_unlocked();} uint32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
int64_t nextlong(void){ char c=getchar_unlocked(); while(c!='-'&&(c<'0'||'9'<c)) c=getchar_unlocked(); int s=0; if(c=='-'){s=1;c=getchar_unlocked();} uint64_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar_unlocked(); } return s?-x:x; }
uint32_t nextstr(char *s){ char c=getchar_unlocked(); while(c==' '||c=='\n') c=getchar_unlocked(); uint32_t len=0; while(c!=' '&&c!='\n'){ *s++=c; len++; c=getchar_unlocked(); } *s='\0'; return len; }

int edge[200000][3];

int qmem[400000][2];

int (*e[100000])[2];
int el[100000];

bool b[100000];
int p[100000];

bool f(int x){
	for(int i=0; i<el[x]; i++){
		int to=e[x][i][0];
		int d=e[x][i][1];
		if(!b[to]){
			p[to]=p[x]+d;
			b[to]=true;
			if(!f(to)) return false;
		}else if(p[to]!=p[x]+d){
			return false;
		}
	}
	return true;
}

int main(void){
	int n=nextint();
	int m=nextint();
	for(int i=0; i<m; i++){
		int A=nextint()-1;
		int B=nextint()-1;
		int C=nextint();
		edge[i][0]=A;
		edge[i][1]=B;
		edge[i][2]=C;
		el[A]++;
		el[B]++;
	}
	int ne=0;
	for(int i=0; i<n; i++){
		e[i]=&qmem[ne];
		ne+=el[i];
		el[i]=0;
	}
	for(int i=0; i<m; i++){
		int A=edge[i][0];
		int B=edge[i][1];
		int C=edge[i][2];
		e[A][el[A]][0]=B;
		e[A][el[A]][1]=C;
		el[A]++;
		e[B][el[B]][0]=A;
		e[B][el[B]][1]=-C;
		el[B]++;
	}
	bool flag=true;
	for(int i=0; i<n; i++){
		if(!b[i] && !f(i)) flag=false;
	}
	puts(flag?"Yes":"No");
}