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<iomanip>
#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int n,m,S,c[55],d[55];
long long dis[55][2505],ans[55],minn[55];
struct E{
	int to,time,cost;
};
struct Node{
	int now,money;
	long long dis;
	bool operator <(const Node yy) const {
		return dis==yy.dis?money>yy.money:dis>yy.dis;
	}
};
vector<E> g[50];
int main(){
	cin>>n>>m>>S;
	S=min(S,2500); 
	for(int i=1;i<=m;i++){
		int x,y,z,w;
		cin>>x>>y>>z>>w;
		g[x].push_back({y,w,z});
		g[y].push_back({x,w,z});
	}
	for(int i=1;i<=n;i++)cin>>c[i]>>d[i],c[i]=min(c[i],2500);
	memset(dis,0x3f,sizeof(dis));
	memset(ans,0x3f,sizeof(ans));
	memset(minn,0x3f,sizeof(minn));
	dis[1][S]=0,minn[1]=-1ll*S*c[1];
	priority_queue<Node> q;
	q.push({1,S,0}); 
	while(!q.empty()){
		Node t=q.top();
		q.pop();
		if(t.dis!=dis[t.now][t.money])continue;
		for(E v:g[t.now]){
			int y=v.to;
			if(t.money-v.cost>=0){
				if(dis[y][t.money-v.cost]>t.dis+v.time){
					dis[y][t.money-v.cost]=t.dis+v.time;
					q.push({y,t.money-v.cost,t.dis+v.time});
					ans[y]=min(ans[y],t.dis+v.time);
				}
			}
		}
		for(int i=t.money+c[t.now],j=1,ii;i<=2500+t.money;i+=c[t.now],j++){
			//if(i%1000==0)cout<<i<<' '<<t.now<<endl;
			if(i>2500)ii=2500;
			else ii=i;
			//if(ii%1000==0)cout<<dis[t.now][ii]<<' '<<t.dis+1ll*j*d[t.now]<<' '<<i<<' '<<t.now<<endl;
			if(dis[t.now][ii]>t.dis+1ll*j*d[t.now]){
				dis[t.now][ii]=t.dis+1ll*j*d[t.now];
				q.push({t.now,ii,dis[t.now][ii]});
			}
			else break;
		}
	}
	for(int i=2;i<=n;i++)cout<<ans[i]<<endl;
    return 0;
}