There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes. You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin. The transaction takes D_i minutes for each gold coin you give. You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

* 2 \leq N \leq 50
* N-1 \leq M \leq 100
* 0 \leq S \leq 10^9
* 1 \leq A_i \leq 50
* 1 \leq B_i,C_i,D_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).
* Each city t=2,...,N can be reached from City 1 with some number of railroads.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N


Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Examples

Input

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5


Output

2
14


Input

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


Output

5
5
7


Input

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1


Output

1
9003
14606
16510
16576


Input

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7


Output

1
3
5


Input

2 1 0
1 2 1 1
1 1000000000
1 1


Output

1000000001
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAXN = 50;
const int MAXCHERCHE = 10000;
bool deja[MAXN][MAXCHERCHE];
bool dU[MAXN];
int mini[MAXN];

struct pr_Q
{
	int temps , noeud , arg;
	bool operator < (const pr_Q & autre) const
	{
		return temps > autre.temps;
	}
};

struct arc
{
	int fin , temps , val;
};

struct Source
{
	int gain , temps;
};

vector <arc> adj[MAXN];
int32_t main()
{
	int nbNoeuds , nbChemins , argent;
	cin >> nbNoeuds >> nbChemins >> argent;
	for (int i = 0 ; i < nbChemins ; i++)
	{
		int a,b,c,d;
		cin >> a >> b >> c >> d;
		a--;b--;
		adj[a].push_back({b,d,c});
		adj[b].push_back({a,d,c});
	}
	Source villes[nbNoeuds];
	for (int i = 0 ; i < nbNoeuds ; i++)
	{
		int a,b;
		cin >> a >> b;
		villes[i] = {a,b};
		mini[i] = 1e18;
	}
	priority_queue<pr_Q> Q;
	Q.push({0 , 0 , argent});
	while (!Q.empty())
	{
		pr_Q actu = Q.top(); Q.pop();
		if (actu.arg < 0) continue;
		if (actu.arg >= MAXCHERCHE && dU[actu.noeud])
		{
			continue;
		}
		if (actu.arg < MAXCHERCHE && deja[actu.noeud][actu.arg])
		{
			continue;
		}
		if (actu.arg >= MAXCHERCHE)
		{
			dU[actu.noeud] = true;
		}
		else
		{
			deja[actu.noeud][actu.arg] = true;
		}
		mini[actu.noeud] = min(mini[actu.noeud] , actu.temps);
		for (auto voisin : adj[actu.noeud])
		{
			Q.push({actu.temps + voisin.temps , voisin.fin , actu.arg - voisin.val });
		}
		Q.push({actu.temps + villes[actu.noeud].temps , actu.noeud , actu.arg + villes[actu.noeud].gain});
	}
	for (int i = 1 ; i < nbNoeuds ; i++)
	{
		cout<<mini[i]<<endl;
	}
}