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;
struct Edge{
	int next,to,w,v;
}edge[N<<2];
struct Node{
	long long val;int u,s;
	bool operator < (const Node &b)const{return val>b.val;}
}u;
int fi[N],se,s,s1,c[N],d[N];long long dis[N][N*N];
bool vis[N][N*N];
inline void add_edge(int u,int v,int a,int b){edge[++se]={fi[u],v,a,b},fi[u]=se;}
inline void add(int u,int v,int a,int b){add_edge(u,v,a,b),add_edge(v,u,a,b);}
priority_queue<Node>q;
void dijk(){
	memset(dis,0x1f,sizeof(dis));dis[1][s]=0;
	q.push(Node{0,1,s});
	while(!q.empty()){
		u=q.top();q.pop();
		if(vis[u.u][u.s])continue;
		vis[u.u][u.s]=1;
		if(dis[u.u][min(s1,u.s+c[u.u])]>dis[u.u][u.s]+d[u.u]){
			dis[u.u][min(s1,u.s+c[u.u])]=dis[u.u][u.s]+d[u.u];
			q.push(Node{dis[u.u][min(s1,u.s+c[u.u])],u.u,min(s1,u.s+c[u.u])});
		}
		for(int i=fi[u.u];i;i=edge[i].next){
			int v=edge[i].to;
			if(u.s>=edge[i].w&&dis[v][u.s-edge[i].w]>dis[u.u][u.s]+edge[i].v){
				dis[v][u.s-edge[i].w]=dis[u.u][u.s]+edge[i].v;
				q.push(Node{dis[v][u.s-edge[i].w],v,u.s-edge[i].w});
			}
		}
	} 
}
int main(){
	int n,m,u,v,a,b;scanf("%d%d%d",&n,&m,&s);
	for(int i=0;i<m;i++)scanf("%d%d%d%d",&u,&v,&a,&b),add(u,v,a,b),s1=max(s1,a);
	for(int i=1;i<=n;i++)scanf("%d%d",c+i,d+i);
	s=min(s,s1*=n);
	dijk();
	for(int i=2;i<=n;i++){
		long long ans=0x1f1f1f1f1f1f1f1f;
		for(int j=0;j<=s1;j++)ans=min(ans,dis[i][j]);
		printf("%lld\n",ans);
	}
	return 0;
}