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 fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> P;

const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;

vector<P> g[maxn];
bool vis[maxn];
int f[maxn];

bool dfs(int u,int d) {
    if(vis[u]) return f[u] == d;
    f[u] = d;
    vis[u] = 1;
    for(auto v:g[u]) {
        if(!dfs(v.fi,d + v.se)) return false;
    }
    return true;
}

int main() {
#ifdef CX_TEST
    freopen("E:\\program--GG\\test_in.txt", "r", stdin);
#endif
    int n, m, i, j, u, v, w;
    scanf("%d%d", &n, &m);
    for(i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &w);
        g[u].push_back(P(v, w));
        g[v].push_back(P(u, -w));
    }
    for(i = 1; i <= n; i++) {
        if(!vis[i] && !dfs(i, 0)) return puts("No"),0;
    }
    puts("Yes");
    return 0;
}