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<iostream>
#include<queue>
#include<algorithm>
using namespace std;
int N,M,S;
vector<pair<int,pair<int,int> > >G[50];
long dist[50][6000];
int C[50],D[50];
int main()
{
	cin>>N>>M>>S;
	for(int i=0;i<M;i++)
	{
		int u,v,a,b;cin>>u>>v>>a>>b;
		u--,v--;
		G[u].push_back(make_pair(v,make_pair(a,b)));
		G[v].push_back(make_pair(u,make_pair(a,b)));
	}
	for(int i=0;i<N;i++)cin>>C[i]>>D[i];
	for(int i=0;i<N;i++)for(int j=0;j<6000;j++)dist[i][j]=4e18;
	dist[0][min(S,5999)]=0;
	priority_queue<pair<long,pair<int,int> > >P;
	P.push(make_pair(0,make_pair(0,min(S,5999))));
	while(!P.empty())
	{
		int u=P.top().second.first;
		int rest=P.top().second.second;
		long c=-P.top().first;
		P.pop();
		if(dist[u][rest]<c)continue;
		{
			int nr=min(rest+C[u],5999);
			long nc=c+D[u];
			if(dist[u][nr]>nc)
			{
				dist[u][nr]=nc;
				P.push(make_pair(-nc,make_pair(u,nr)));
			}
		}
		for(pair<int,pair<int,int> >e:G[u])
		{
			int v=e.first;
			int nr=rest-e.second.first;
			long nc=c+e.second.second;
			if(nr<0)continue;
			if(dist[v][nr]>nc)
			{
				dist[v][nr]=nc;
				P.push(make_pair(-nc,make_pair(v,nr)));
			}
		}
	}
	for(int t=1;t<N;t++)
	{
		long ans=4e18;
		for(int i=0;i<6000;i++)ans=min(ans,dist[t][i]);
		cout<<ans<<endl;
	}
}
