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 <iostream>
#include <cstdio>
#include <vector>
#include <cstdlib>

using namespace std;

//const int SP = 2100000000;
int n, m;
int l, r, d;
int a[100005];
int b[100005];
bool ans = true;
vector<int> vx[100005], wt[100005];

void dfs(int p, int q)
{
    if(b[p])
    {
        if(a[p] != q) printf("No"), exit(0);
        return;
    }
    b[p] = true;
    a[p] = q;
    for(int i = 0; i < vx[p].size(); i++)
    {
        dfs(vx[p][i], wt[p][i] + q);
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++)
    {
        scanf("%d%d%d", &l, &r, &d);
        vx[l].push_back(r);
        vx[r].push_back(l);
        wt[l].push_back(d);
        wt[r].push_back(-d);
    }
    for(int i = 1; i <= n; i++)
    {
        if(!b[i]) dfs(i, 0);
    }
    printf("Yes");
    return 0;
}