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;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<int,pii> piii;
typedef pair<ll,pii> plii;
#define mkp make_pair
#define fi first
#define se second
int n,m,S;
ll d[53][5003],ans;
vector<piii>g[53];
int C[53],D[53];
bool vis[53][5003];
priority_queue <plii,vector<plii>,greater<plii> >pq;
void GO(int a,int b,ll c){
	if (c<d[a][b]){
		d[a][b]=c;
		pq.push(mkp(c,mkp(a,b)));
	}
}
int main(){
	cin>>n>>m>>S;
	S=min(S,5000);
	for (int i=0;i<m;i++){
		int u,v,a,b;cin>>u>>v>>a>>b;
		g[u].push_back(mkp(v,mkp(a,b)));
		g[v].push_back(mkp(u,mkp(a,b)));
	}
	for (int i=1;i<=n;i++) cin>>C[i]>>D[i];
	for (int i=0;i<53;i++) for(int j=0;j<5003;j++)d[i][j]=1e18;
	d[1][S]=0;
	pq.push(mkp(0,mkp(1,S)));
	while (!pq.empty()){
		plii x=pq.top();pq.pop();
		if (vis[x.se.fi][x.se.se]) continue;vis[x.se.fi][x.se.se]=1;
		GO(x.se.fi,min(5000,x.se.se+C[x.se.fi]),x.fi+D[x.se.fi]);
		for (int i=0;i<g[x.se.fi].size();i++)
			if (g[x.se.fi][i].se.fi<=x.se.se)
				GO(g[x.se.fi][i].fi,x.se.se-g[x.se.fi][i].se.fi,x.fi+g[x.se.fi][i].se.se);
	}
	for (int i=2;i<=n;i++){
		ans=1e18;
		for (int j=0;j<=5000;j++) ans=min(ans,d[i][j]);
		cout<<ans<<"\n";
	}
}