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;

int N, M, ap[100009];
long long h[100009];
vector < pair < int, int > > v[100009];

void dfs (int nod)
{
    ap[nod] = 1;
    for (auto it : v[nod])
        if (ap[it.first] == 0)
            h[it.first] = h[nod] + it.second, dfs (it.first);
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &M);
while (M --)
{
    int x, y, d;
    scanf ("%d %d %d", &x, &y, &d);
    v[x].push_back ({y, d});
    v[y].push_back ({x, -d});
}
for (int i=1; i<=N; i++)
    if (ap[i] == 0)
        h[i] = 0, dfs (i);
for (int i=1; i<=N; i++)
    for (auto it : v[i])
        if (h[it.first] != h[i] + it.second)
        {
            printf ("No\n");
            return 0;
        }
printf ("Yes\n");
return 0;
}