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<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF (1LL<<60)
#define MOD1 1000000007

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a=c;}/*swap 交換*/
ll max2(ll a,ll b){return a>=b?a:b;}/*2つのうち大きい数を返す*/ 
ll min2(ll a,ll b){return a>=b?b:a;}/*2つのうち小さい数を返す*/ 
ll ABS(ll a){return a>=0?a:(-a);}/*絶対値*/ 
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

/*-------------------------------------------------------------*/
// Weighted Union Find
typedef struct{
    ll num; //number
    ll par; //parent
    ll hei; //height
    ll wdif;
    ll wei;
}WUF;
void WUFmake(ll n, WUF b[n]){for(ll i=0;i<n;i++){
    b[i].par = i;
    b[i].num = i;
    b[i].hei = 1;
    b[i].wdif = 0;
    b[i].wei = 0;
    }
}
ll WUFroot(ll n, ll a, WUF b[n]){
    ll c= a==b[a].par ? a : WUFroot(n,b[a].par,b) ;
    b[a].wei = b[b[a].par].wei + b[a].wdif;
    return c;
}
// b[x]+z=b[y]
void WUFunite(ll n, ll x, ll y, ll z, WUF b[n]){
    ll xr = WUFroot(n,x,b), yr = WUFroot(n,y,b);
    if(b[xr].hei > b[yr].hei){ b[yr].par = xr;  b[yr].wdif = b[x].wei+z-b[y].wei;
    }else{ b[xr].par = yr;  b[yr].hei = max2(b[yr].hei,b[xr].hei+1);  b[xr].wdif = b[y].wei-b[x].wei-z;}
}
ll WUFsame(ll n, ll x, ll y, WUF b[n]){ return WUFroot(n,x,b) == WUFroot(n,y,b) ? 1 : 0; }



int main(void){
    ll n,m;
    scanf("%lld%lld",&n,&m);
    // if(m==0){
    //     printf("Yes\n");
    //     return 0;
    // }
    ll l[m],r[m],d[m];
    WUF tree[n];
    WUFmake(n,tree);
    rep(i,0,m){
        scanf("%lld%lld%lld",&l[i],&r[i],&d[i]);
    }
    rep(i,0,m){
        if(WUFsame(n,l[i]-1,r[i]-1,tree)){
            if(tree[r[i]-1].wei - tree[l[i]-1].wei != d[i]){
                printf("No\n");
                return 0;
            }
        }else{
            WUFunite(n,l[i]-1,r[i]-1,d[i],tree);
        }
    }
    printf("Yes\n");
    return 0;
}