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>
#define ll long long int
#define mod 1000000007
#define pii pair<int, int>
#define fr(n) for (int i = 0; i < n; i++)
#define fr1(n) for (int i = 1; i <= n; i++)
using namespace std;

vector<pii> g[100005];
int d[100005];
bool b[100005];

void dfs(int x) {
    for (pii i : g[x]) {
        if (!b[i.first]) {
            b[i.first] = 1;
            d[i.first] = d[x] + i.second;
            dfs(i.first);
        } else {
            if (d[i.first] != d[x] + i.second) {
                cout << "No";
                exit(0);
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    int n, m, u, v, w;
    cin >> n >> m;
    fr(m) cin >> u >> v >> w, g[u].push_back({v, w}), g[v].push_back({u, -w});
    fr1(n) if (!b[i]) b[i] = 1, dfs(i);
    cout << "Yes";
}