Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
using ll = long long;
using vec = vector<ll>;
using vect = vector<double>;
using Graph = vector<vector<ll>>;
#define loop(i, n) for (ll i = 0; i < n; i++)
#define Loop(i, m, n) for (ll i = m; i < n; i++)
#define pool(i, n) for (ll i = n; i >= 0; i--)
#define Pool(i, m, n) for (ll i = n; i >= m; i--)
#define mod 1000000007ll
//#define mod 998244353ll
#define flagcount __builtin_popcount
#define flag(x) (1ll << x)
#define flagadd(bit, x) bit |= flag(x)
#define flagpop(bit, x) bit &= ~flag(x)
#define flagon(bit, i) bit &flag(i)
#define flagoff(bit, i) !(bit & (1ll << i))
#define all(v) v.begin(), v.end()
#define low2way(v, x) lower_bound(all(v), x)
#define high2way(v, x) upper_bound(all(v), x)
#define idx_lower(v, x) (distance(v.begin(), low2way(v, x)))  //配列vでx未満の要素数を返す
#define idx_upper(v, x) (distance(v.begin(), high2way(v, x))) //配列vでx以下の要素数を返す
#define idx_lower2(v, x) (v.size() - idx_lower(v, x))         //配列vでx以上の要素数を返す
#define idx_upper2(v, x) (v.size() - idx_upper(v, x))         //配列vでxより大きい要素の数を返す
#define putout(a) cout << a << endl
#define Sum(v) accumulate(all(v), 0ll)
#define gcd(x, y) __gcd(x, y)
ll ctoi(char c)
{
    if (c >= '0' && c <= '9')
    {
        return c - '0';
    }
    return 0;
}
template <typename T>
T lcm(T x, T y)
{
    T z = gcd(x, y);
    return x * y / z;
}
template <typename T>
bool primejudge(T n)
{
    if (n < 2)
        return false;
    else if (n == 2)
        return true;
    else if (n % 2 == 0)
        return false;
    double sqrtn = sqrt(n);
    for (T i = 3; i < sqrtn + 1; i++)
    {
        if (n % i == 0)
        {
            return false;
        }
        i++;
    }
    return true;
}
//場合によって使い分ける
//const ll dx[4]={1,0,-1,0};
//const ll dy[4]={0,1,0,-1};
const ll dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//2次元配列の宣言
//vector<vector<ll>> field(h, vector<ll>(w));

struct weight_union_find
{
    vector<long long> par;      //親の番号　
    vector<long long> rank;     //木の深さ(根のランクは0)
    vector<long long> siz;      //要素xが根なら木の頂点数を格納する
    vector<long long> par_diff; //親と子のコストの差
    //初期化子リストを用いた初期化
    weight_union_find(long long N) : par(N), rank(N), siz(N), par_diff(N)
    {
        for (long long i = 0; i < N; i++)
        {
            par[i] = i;
            rank[i] = 0;
            siz[i] = 1;
            par_diff[i] = 0;
        }
    }
    //要素xが所属する木の根を再帰的に発見する
    long long root(long long x)
    {
        if (par[x] == x)
            return x;
        else
        {
            //経路圧縮+累積和
            par_diff[x] += par_diff[par[x]];
            return par[x] = root(par[x]);
        }
    }
    //経路圧縮+コストを返す
    long long weight(long long x)
    {
        root(x);
        return par_diff[x];
    }
    //yのコスト-xのコスト
    long long diff(long long x, long long y)
    {
        return weight(y) - weight(x);
    }
    //weight(y)=weight(x)+wとなるように木を併合
    void unite(long long x, long long y, long long w)
    {
        long long rx = root(x);
        long long ry = root(y);
        w += weight(x);
        w -= weight(y);
        if (rx == ry)
            return; //同じ木に属してたらそのまま
        if (rank[rx] < rank[ry])
        {
            swap(rx, ry);
            siz[ry] = siz[rx] + siz[ry];
            w *= (-1); //符号反転
        }
        else
        {
            siz[rx] = siz[rx] + siz[ry];
        }
        if (rank[rx] == rank[ry])
            rank[rx]++;
        par[ry] = rx;
        par_diff[ry] = w;
    }
    //要素xが属する木と要素yが属する木が同じならtrueを返す
    bool same(long long x, long long y)
    {
        return root(x) == root(y);
    }
    //要素xが属する木の頂点数を返す
    long long size(long long x)
    {
        return siz[root(x)];
    }
};
/*
この重み付きunion_find木で出来ること
例)weight_union_find tree(N);などで宣言(N:頂点数)
tree.same(x,y):要素xと要素yが同じグループに属するか判定
tree.unite(x,y,w):weight(y)=weight(x)+wとなるように木を併合
tree.diff(x,y):weight(y)-weight(x)を返す
tree.size(x):要素xが属する木の頂点数を返す
*/
int main()
{
    cout << fixed << setprecision(30);
    ll N, M;
    cin >> N >> M;
    vec l(M), r(M), d(M);
    weight_union_find tree(N);
    loop(i, M)
    {
        cin >> l[i] >> r[i] >> d[i];
        l[i]--;
        r[i]--;
        if (tree.same(l[i], r[i]))
        {
            if (tree.diff(l[i], r[i]) != d[i])
            {
                putout("No");
                return 0;
            }
        }
        else
            tree.unite(l[i], r[i], d[i]);
    }
    putout("Yes");
    return 0;
}