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 inf 999999999999999999
#define int long long 
#define N 666666
int dis[66][2600]; 
int vis[66][2600]; 
int n,m,s;

int C[N],D[N];
//*********************建图 
int head[N];
int cnt;
struct edge{
	int u;
	int v;
	int a;
	int b;
	int next;
}e[N];
void addedge(int u,int v,int a,int b){
	e[++cnt].v=v;
	e[cnt].a=a;
	e[cnt].b=b;
	e[cnt].next=head[u];
	head[u]=cnt;
}

void init(){
	for(int i=0;i<=n;i++)
		for(int j=0;j<=2555;j++)
			dis[i][j]=inf;
	return ;
}

void dij(){
	dis[1][s]=0;
	 queue<pair<int,int> > q;
	 q.push(make_pair(1,s));
	 while(!q.empty()){
	 	int u=q.front().first;
	 	int w=q.front().second;
	 	q.pop();
	 	for(int i=head[u];i;i=e[i].next){
	 		int v=e[i].v;
	 		int a=e[i].a;
	 		int b=e[i].b;
	 		if(w-a>=0&&dis[v][w-a]>dis[u][w]+b){
	 			dis[v][w-a]=min(dis[v][w-a],dis[u][w]+b);
	 			q.push(make_pair(v,w-a));
			}
		}
		if(dis[u][w]+D[u]<dis[u][min(2500*1ll,w+C[u])]){
			dis[u][min(2500*1ll,w+C[u])]=dis[u][w]+D[u];
			q.push(make_pair(u,min(2500*1ll,w+C[u])));
		}
	 }
}
signed main(){
	
	cin>>n>>m>>s;
	s=min(s,2500*1ll);
	init();
	for(int i=1;i<=m;i++){
		int a,b,c,d;
		scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
		addedge(a,b,c,d);
		addedge(b,a,c,d);
	}
	for(int i=1;i<=n;i++) scanf("%lld%lld",&C[i],&D[i]);
	dij();
	
	for(int i=2;i<=n;i++){
		int res=inf;
		for(int j=0;j<=2500;j++) res=min(res,dis[i][j]);
		printf("%lld\n",res);
	} 
	return 0;
} 