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>
#define int long long
using namespace std;
const int N=59,M=109;
struct edge{int to,nxt,a,w;}e[M*2]; int hd[N],tot;
void add(int u,int v,int a,int w){e[++tot]=(edge){v,hd[u],a,w};hd[u]=tot;}
int n,m,s; int maxs; int c[N],d[N]; int dis[N][N*N],ans[N];
struct node{
	int u,sil,f;
	bool operator < (const node b)const{return f>b.f;}
	bool operator > (const node b)const{return f<b.f;}
};
void dijkstra(){
	memset(dis,0x3f,sizeof(dis)); memset(ans,0x3f,sizeof(ans));
	priority_queue<node>q; q.push((node){1,s}); dis[1][s]=0;
	while(!q.empty()){
		int u=q.top().u,sil=q.top().sil; q.pop();
		ans[u]=min(ans[u],dis[u][sil]);
		for(int i=hd[u],v;i;i=e[i].nxt){
			if(e[i].a<=sil&&dis[v=e[i].to][sil-e[i].a]>dis[u][sil]+e[i].w)
				q.push((node){v,sil-e[i].a,dis[v][sil-e[i].a]=dis[u][sil]+e[i].w});
		}
		if(sil<maxs&&dis[u][min(sil+c[u],maxs)]>dis[u][sil]+d[u])
			q.push((node){u,min(maxs,sil+c[u]),dis[u][min(sil+c[u],maxs)]=dis[u][sil]+d[u]});
	}
}
signed main(){
	scanf("%lld%lld%lld",&n,&m,&s);
	for(int i=1,u,v,a,w;i<=m;i++)
		scanf("%lld%lld%lld%lld",&u,&v,&a,&w),add(u,v,a,w),add(v,u,a,w),maxs+=a;
	s=min(s,maxs);
	for(int i=1;i<=n;i++) 
		scanf("%lld%lld",&c[i],&d[i]);
	dijkstra();
	for(int i=2;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}