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 rep(i, n) for(int i=0; i<n; ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
#define ll long long int

struct Edge{
	int to, a, b;
	Edge(int to, int a, int b):to(to),a(a),b(b){}
};

struct Data{
	int v, s;
	ll x;
	Data(int v, int s, ll x):v(v), s(s), x(x){}
	bool operator<(const Data& a) const {
		return x > a.x;
	}
};

#define S_MAX 2500

int main(){
	int n, m, s;
	cin>>n>>m>>s;
	vector<Edge> e[50];
	rep(i, m){
		int u, v, a, b;
		cin>>u>>v>>a>>b; u--; v--;
		e[u].push_back(Edge(v, a, b));
		e[v].push_back(Edge(u, a, b));
	}
	int c[50], d[50];
	rep(i, n) cin>>c[i]>>d[i];

	priority_queue<Data> q;

	ll mm[50][S_MAX];
	rep(i, 50) rep(j, S_MAX) mm[i][j]=LLONG_MAX;
	s=min(s, S_MAX-1);
	mm[0][s]=0;
	q.push(Data(0, s, 0));

	while(!q.empty()){
		Data da=q.top(); q.pop();
		if(da.x>mm[da.v][da.s]) continue;

		repit(it, e[da.v]){
			int ns=da.s-(it->a);
			ll nx=da.x+(it->b);
			if(ns>=0 && nx<mm[it->to][ns]){
				mm[it->to][ns]=nx;
				q.push(Data(it->to, ns, nx));
			}
		}
		for(ll ns=da.s, nx=da.x; ; ){
			ns+=c[da.v];
			nx+=d[da.v];
			if(ns>=S_MAX) break;
			if(nx < mm[da.v][ns]){
				mm[da.v][ns]=nx;
				q.push(Data(da.v, ns, nx));
			}
			
		}
	}

	for(int i=1; i<n; i++){
		ll mn=LLONG_MAX;
		rep(j, S_MAX) mn=min(mn, mm[i][j]);
		cout<<mn<<endl;
	}
	return 0;
}
