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>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const int MAX_N = 55;
const ll INF = 1e15;
ll C[MAX_N],D[MAX_N],t[MAX_N*MAX_N*50];
priority_queue<P, vector<P>, greater<P> > Q;

void init()
{
	rep(i,MAX_N*MAX_N*50)
	{
		t[i] = INF;
	}
	return;
}

struct edge
{
	ll to, cost, time;
};

vector<edge> G[MAX_N];

struct edge2
{
	ll to, time;
};

vector<edge2> G2[MAX_N*MAX_N*50];

int main()
{
	ll n,m,s;
	cin >> n >> m >> s;
	rep(i,m)
	{
		ll u,v,a,b;
		cin >> u >> v >> a >> b;
		u--; v--;
		/*
		edge e;
		e.to = v; e.cost = a, e.time = b;
		G[u].push_back(e);
		e.to = v;
		G[v].push_back(e);
		*/
		int temp = a;
		while(temp <= MAX_N*50)
		{
			edge2 e2;
			e2.to = (temp-a)*n+v; e2.time = b;
			G2[temp*n+u].push_back(e2);
			e2.to = (temp-a)*n+u;
			G2[temp*n+v].push_back(e2);
			temp++;
		}
	}
	rep(i,n)
	{
		cin >> C[i] >> D[i];
		rep(j,MAX_N*50)
		{
			edge2 e2;
			e2.time = D[i];
			if (j+C[i] > MAX_N*50) e2.to = n * MAX_N * 50 + i;
			else e2.to = n*(j+C[i]) + i;
			G2[i+j*n].push_back(e2);
		}
	}

	init();
	s = min(s,ll(MAX_N*50));
	t[s*n] = 0;
	Q.push(P(0,s*n));
	while(!Q.empty())
	{
		P p = Q.top(); Q.pop();
		ll v = p.second;
		if (t[v] < p.first) continue;
		rep(i,G2[v].size())
		{
			edge2 e = G2[v][i];
			if (t[e.to] > t[v] + e.time)
			{
				t[e.to] = t[v] + e.time;
				Q.push(P(t[e.to], e.to));
			}
		}
	}

	rep(i,n-1)
	{
		ll ans = INF;
		rep(j,MAX_N*50+1)
		{
			ans = min(ans,t[j*n+i+1]);
		}
		cout << ans << endl;
	}
	return 0;
}