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>
#define N 100005
using namespace std;

int n, m, l[N * 2], r, d, a[N];
bool v[N];
vector<int> ve[N], vd[N];

void f(int p) {
	int i;
	for (i = 0; i < ve[p].size(); i++) {
		if (!v[ve[p][i]]) v[ve[p][i]] = 1, a[ve[p][i]] = a[p] + vd[p][i], f(ve[p][i]);
		else if (a[p] + vd[p][i] != a[ve[p][i]]) {cout << "No"; exit(0);}
	}
}

int main()
{
	int i;
	cin >> n >> m;
	for (i = 0; i < m; i++) {
		scanf("%d %d %d", &l[i], &r, &d);
		ve[l[i]].push_back(r); vd[l[i]].push_back(d);
		ve[r].push_back(l[i]); vd[r].push_back(-d);
	}
	for (i = 0; i < m; i++) if (!v[l[i]]) v[l[i]] = 1, f(l[i]);
    cout << "Yes" << endl;
    return 0;
}