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>
 
using namespace std;

typedef pair<long long,long long> ii;
typedef pair<long long,ii> iii;

long long me[50][60*50],c[50],d[50];
long long res[50];

vector < vector <iii> > mv;

int main(){
	
	cin.tie(0); ios_base::sync_with_stdio(0);
	long long n,m,s,u,v,a,b;
	cin>>n>>m>>s;
	mv.resize(n);
	for(int i=0;i<m;i++){
		cin>>u>>v>>a>>b;
		u--;v--;
		mv[u].push_back(iii(v,ii(a,b)));
		mv[v].push_back(iii(u,ii(a,b)));
	}
	for(int i=0;i<n;i++){
		cin>>c[i]>>d[i];
		for(int j=0;j<=n*50;j++)me[i][j] = 1000000000000000;
		res[i] = 1000000000000000;
	}
	if(s>n*50)s=n*50;
	me[0][s] = 0;
	res[0] = 0;
	priority_queue <iii> mq;
	mq.push(iii(0,ii(0,s)));
	long long t,ac,ns,ts;
	while(!mq.empty()){
		iii top = mq.top();
		mq.pop();
		t = -top.first;
		ac=top.second.first;
		s = top.second.second;
		//if(ac==0)cout<<ac<<' '<<s<<' '<<t<<' '<<me[ac][s]<<endl;
		if(me[ac][s]==t){
			if(t<res[ac])res[ac] = t;
			if(s<n*50){
				ns = s +c[ac];
				if(ns>n*50)ns = n*50;
				ts = t +d[ac];
				if(ts<me[ac][ns]){
					me[ac][ns] = ts;
					mq.push(iii(-ts,ii(ac,ns)));
				}
			}
			for(int i=0;i<mv[ac].size();i++){
				v = mv[ac][i].first;
				a = mv[ac][i].second.first;
				b = mv[ac][i].second.second;
				ts = t + b;
				if(a<=s&&me[v][s-a]>ts){
					me[v][s-a]=ts;
					mq.push(iii(-ts,ii(v,s-a)));
				}
			}
		}
	}
	for(int i=1;i<n;i++)cout<<res[i]<<endl;
	
	return 0;
}