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;
typedef long long ll;
const int N = 100100;
int n , m ;
int vis[N];
ll dis[N];
vector<pair<int,ll>>adj[N];

bool dfs(int u,ll d){

if( vis[u] )return dis[u]==d;
vis[u]=1;
dis[u]=d;
for(auto p:adj[u]){
    int v = p.first;
    ll dx = p.second;
    if( !dfs(v,d+dx) )
        return 0;
}
return 1;
}

int main(){

scanf("%d %d",&n,&m);
while( m-- ){
    int i,j;
    ll d;
    scanf("%d %d %lld",&i,&j,&d);
    adj[i].push_back({j,d});
    adj[j].push_back({i,-d});
}

for(int i=1;i<=n;i++)
    if( !vis[i] )
if( !dfs(i,0) ){
    printf("No\n");
    return 0;
}

printf("Yes\n");
return 0;
}