There are n cities and m bidirectional roads in the country. The roads in the country form an undirected weighted graph. The graph is not guaranteed to be connected. Each road has it's own parameter w. You can travel through the roads, but the government made a new law: you can only go through two roads at a time (go from city a to city b and then from city b to city c) and you will have to pay (w_{ab} + w_{bc})^2 money to go through those roads. Find out whether it is possible to travel from city 1 to every other city t and what's the minimum amount of money you need to get from 1 to t.

Input

First line contains two integers n, m (2 ≤ n ≤ 10^5, 1 ≤ m ≤ min((n ⋅ (n - 1))/(2), 2 ⋅ 10^5)).

Next m lines each contain three integers v_i, u_i, w_i (1 ≤ v_i, u_i ≤ n, 1 ≤ w_i ≤ 50, u_i ≠ v_i). It's guaranteed that there are no multiple edges, i.e. for any edge (u_i, v_i) there are no other edges (u_i, v_i) or (v_i, u_i).

Output

For every city t print one integer. If there is no correct path between 1 and t output -1. Otherwise print out the minimum amount of money needed to travel from 1 to t.

Examples

Input


5 6
1 2 3
2 3 4
3 4 5
4 5 6
1 5 1
2 4 2


Output


0 98 49 25 114 

Input


3 2
1 2 1
2 3 2


Output


0 -1 9 

Note

The graph in the first example looks like this.

<image>

In the second example the path from 1 to 3 goes through 2, so the resulting payment is (1 + 2)^2 = 9.

<image>
# include <bits/stdc++.h>
using namespace std;

#define ll          long long
#define int          ll
#define pb          push_back
#define ppb         pop_back
#define eb          emplace_back
#define endl		'\n'
#define max(a,b)	((a)>(b)?(a):(b))
#define min(a,b)	((a)<(b)?(a):(b))
#define vi          vector<int>
#define vii         vector<pair<int, int> >
#define mp          make_pair
#define mem1(a)     memset((a),-1,sizeof(a))
#define mem0(a)     memset((a),0,sizeof(a))
#define memf(a)     memset((a),false,sizeof(a))
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define gcd(a,b)    __gcd((a),(b))
#define lcm(a,b)    (a)*((b)/gcd((a),(b)))
#define DECIMAL(n)  cout << fixed ; cout << setprecision(n);
#define FAST        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mi          map<int,int>
#define sz(x)       (ll)(x).size()
#define rep(i,a,b)  for(int i=(a);i<(b);i++)
#define pii         pair<int,int>
#define INF         1000000000000000000

vector<pii> adj[100001];
int ans[100001];
int d[100001];
int p[100001];

int solve()
{
	int n;
	cin >> n;
	int m;
	cin >> m;
	rep(i, 0, m)
	{
		int u, v, w;
		cin >> u >> v >> w;
		adj[u].pb({v, w});
		adj[v].pb({u, w});
	}
	rep(i, 1, n + 1)
	{
		sort(all(adj[i]));
	}
	ans[1] = 0;
	d[1] = INF;
	rep(i, 2, n + 1)
	{
		ans[i] = INF;
		d[i] = INF;
		p[i] = -1;
	}
	set<pair<int, int>> q;
	// set<int> abc;
	q.insert({0, 1});
	bool visited[100001];
	memf(visited);
	while (!q.empty())
	{
		vi abc;
		auto it = q.begin();
		// set<int> abc;
		int w = q.begin()->F;
		int v = q.begin()->S;
		// cout << v << endl;
		// int w1=it.S.S;
		visited[v] = true;
		q.erase(it);
		for (auto u : adj[v])
		{
			if (d[u.F] > u.S)
			{
				d[u.F] = u.S;
				p[u.F] = v;
				abc.pb(u.F);
			}
		}
		for (auto u : abc)
		{
			for (auto w : adj[u])
			{
				int node = w.F;
				int weight = w.S;
				if (!visited[node])
				{
					if (ans[node] > (ans[p[u]] + (d[u] + weight) * (d[u] + weight)))
					{
						q.erase({ans[node], node});
						ans[node] = (ans[p[u]] + (d[u] + weight) * (d[u] + weight));
						q.insert({ans[node], node});
					}
				}
			}
		}

	}
	rep(i, 1, n + 1)
	{
		if (ans[i] < INF)
			cout << ans[i] << " ";
		else
			cout << "-1 ";

	}

	return 0;
}
signed main()
{
	FAST
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	int t = 1;
	// cin>>t;
	while (t)
	{
		solve();
		t--;
	}
	return 0;
}