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;
const int N=55;
typedef long long LL;
struct edge{
	int to,next,len,cost;
}e[N*4]; 
int n,m,tot,c[N],d[N],s,mx,bk[N],first[N],mxi;
LL dp[N][N*N*2];
void add(int u,int v,int a,int b){
	e[++tot].next=first[u],first[u]=tot;
	e[tot].len=b,e[tot].to=v;e[tot].cost=a;
}

void spfa(){
	queue<int>q;
	q.push(1);bk[1]=1;
	while(!q.empty()){
		int now=q.front();q.pop();
		for(int i=0;i<=mx;i++)
			for(int j=1;j*c[now]<=i;j++){
				dp[now][i]=min(dp[now][i],dp[now][i-j*c[now]]+1ll*j*d[now]);
				//cout<<now<<" "<<dp[now][i]<<endl;
			}
		for(int i=first[now];i;i=e[i].next){
			int u=e[i].to;
			for(int j=0;j<=mx;j++){
				if(e[i].cost<=j){
					if(dp[u][j-e[i].cost]>dp[now][j]+e[i].len){
						dp[u][j-e[i].cost]=dp[now][j]+e[i].len;
						if(!bk[u])bk[u]=1,q.push(u);
					}
				}
			}
		}
		bk[now]=0;
	}
}

int main(){
	scanf("%d%d%d",&n,&m,&s);
	for(int i=1,u,v,a,b;i<=m;i++){
		scanf("%d%d%d%d",&u,&v,&a,&b);
		add(u,v,a,b);add(v,u,a,b);
		if(a>mxi)mxi=a;
	}
	for(int i=1;i<=n;i++){
		scanf("%d%d",&c[i],&d[i]);
		
	}
	if(s>1000000)mx=mxi*n*2;
	else{
		mx=n*n*2+1;
	}
	if(s>mx)s=mx;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=mx;j++)
			dp[i][j]=1e17;
	for(int i=0;i<=s;i++)
		dp[1][i]=0;
	spfa();
	for(int i=2;i<=n;i++){
		LL tmp=1e17;
		for(int j=0;j<=mx;j++)
			if(dp[i][j]<tmp)tmp=dp[i][j];
		printf("%lld\n",tmp);
	}
}