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 ll long long
const ll inf=1e18+69;
int n, m, s;
vector<pair<int, pair<int, ll> > >adj[55];
pair<ll, ll>cd[55];
ll dist[55][5005];
bool cmp(pair<ll, pair<ll, int> >a, pair<ll, pair<ll, int> >b){
	if(a.first==b.first) return a.second.first>b.second.first;
	return a.first>b.first;
}
int main(){
	cin>>n>>m>>s;
	for(int i=0; i<m; i++){
		int u, v, a; ll b; cin>>u>>v>>a>>b;
		adj[u].push_back({v,{a,b}});
		adj[v].push_back({u,{a,b}});
	}
	if(s>=5005) s=5004;
	for(int i=1; i<=n; i++)
		cin>>cd[i].first>>cd[i].second;
	priority_queue<pair<ll, pair<ll, int> >, vector<pair<ll, pair<ll, int> > >, function<bool(pair<ll, pair<ll, int> >, pair<ll, pair<ll, int> >)> >pq(cmp);
	for(int i=1; i<=n; i++)
		for(int j=0; j<5005; j++)
			dist[i][j]=inf;
	dist[1][s]=0ll;
	pq.push({0,{s,1}});
	for(;!pq.empty();){
		ll d=pq.top().first, sc=pq.top().second.first, nw=pq.top().second.second;
		pq.pop();
		if(dist[nw][sc]>d) continue;
		if(sc+cd[nw].first<5005 && dist[nw][sc+cd[nw].first]>dist[nw][sc]+cd[nw].second){
			dist[nw][sc+cd[nw].first]=dist[nw][sc]+cd[nw].second;
			pq.push({dist[nw][sc+cd[nw].first], {sc+cd[nw].first, nw}});
		}
		for(auto i:adj[nw]){
			int nx=i.first, cst=i.second.first; ll tm=i.second.second;
			if(sc<cst) continue;
			if(dist[nx][sc-cst]>dist[nw][sc]+tm){
				dist[nx][sc-cst]=dist[nw][sc]+tm;
				pq.push({dist[nx][sc-cst], {sc-cst, nx}});
			}
		}
	}
	for(int i=2; i<=n; i++){
		ll ans=inf;
		for(int j=0; j<5005; j++)
			ans=min(ans, dist[i][j]);
		cout<<ans<<endl;
	}
	return 0;
}