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;

#define MAX_N 100001

vector<pair<int, int64_t> > adj[MAX_N];
int64_t dis[MAX_N];

void dfs(int v, int64_t d)
{
    if (dis[v]) {
        if (dis[v] != d) {
            printf("No\n");
            exit(0);
        }
        return;
    }
    dis[v] = d;
    for (int i = 0; i < adj[v].size(); i++) {
        int u = adj[v][i].first;
        int64_t w = adj[v][i].second;
        dfs(u, d + w);
    }
}

int main(void)
{
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++) {
        int l, r, d;
        scanf("%d%d%d", &l, &r, &d);
        adj[l].push_back(make_pair(r, d));
        adj[r].push_back(make_pair(l, -d));
    }
    for (int i = 1; i <= n; i++)
        if (!dis[i]) dfs(i, 0xffff);
    printf("Yes\n");
    return 0;
}