Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

#define pb push_back
#define mp make_pair
#define st first
#define nd second
typedef pair < int , int > pp;
const int mod = 1e9 + 7;
const int N   = 2e5 + 5;

vector < pp > V[N];
int T[N],H[N],n,m,i,x,y,z;

void f(int x){
    H[x] = 1;
    for(int y, i=0; i<V[x].size(); i++){
        if(!H[ y = V[x][i].st ]){
            T[y] = T[x] + V[x][i].nd;
            f(y);
        }
        else if(T[y] != T[x] + V[x][i].nd){
            puts("No"); exit(0);
        }
    }
}

signed main(){
    scanf("%d%d",&n,&m);
    for(; m-- ;){
        scanf("%d%d%d",&x,&y,&z);
        V[x].pb(mp(y,z));
        V[y].pb(mp(x,-z));
    }

    for(i=1;i<=n;i++){
        if(H[i]) continue;
        f(i);
    }

    puts("Yes");
    return 0;
}