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;
const int maxn=112345;
typedef pair<int,int> pii;
int n,m,l,r,x,d[maxn],vis[maxn];
vector<pii> G[maxn];
bool dfs(int u,int dep) {
    vis[u]=1;
    d[u]=dep;
    for (int i=0;i<(int)G[u].size();++i) {
        int v=G[u][i].first,w=G[u][i].second;
        if (vis[v]&&d[u]+w!=d[v]) return false;
        if (!vis[v]&&!dfs(v,dep+w)) return false;
    }
    return true;
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;++i) {
        scanf("%d%d%d",&l,&r,&x);
        G[l].push_back(pii(r,x));
        G[r].push_back(pii(l,-x));
    }
    for (int i=1;i<=n;++i) if (!vis[i]&&!dfs(i,0))
        return 0*puts("No");
    return 0*puts("Yes");
}