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>
#include<vector>
#include<queue>
using namespace std;
typedef long long LL;
const LL INF=1e18;
struct node{
	LL d; int u,w;
	node() { }
	node(LL _d,int _u,int _w) { d=_d; u=_u; w=_w; }
	bool operator < (const node &b) const
	{
		return d>b.d;
	}
};
struct edge{
	int v,a,b;
	edge( ) { }
	edge(int _v,int _a,int _b) { v=_v; a=_a; b=_b; }
};
vector<struct edge> e[55];
LL dis[55][2505]; int c[55]; int dd[55];
int main(void)
{
	int n,m,s; scanf("%d%d%d",&n,&m,&s); s=min(s,2500);
	for(int i=1;i<=m;i++)
	{
		int u,v,a,b; scanf("%d%d%d%d",&u,&v,&a,&b);
		e[u].push_back(edge(v,a,b));
		e[v].push_back(edge(u,a,b));
	}
	for(int i=1;i<=n;i++) scanf("%d%d",&c[i],&dd[i]);
	for(int i=1;i<=n;i++) for(int j=0;j<=2500;j++) dis[i][j]=INF;
	dis[1][s]=0;
	priority_queue<struct node> q;
	q.push(node(dis[1][s],1,s));
	while(!q.empty())
	{
		struct node TT=q.top(); q.pop();
		LL d=TT.d; int u=TT.u; int w=TT.w;
		if(d>dis[u][w]) continue;
//		printf("d==%lld,u==%d,w==%d\n",d,u,w);
		
		for(auto EE:e[u])
		{
			int v=EE.v; int a=EE.a; int b=EE.b;
			if(w>=a&&dis[v][w-a]>d+b)	
			{
				dis[v][w-a]=d+b;
				q.push(node(dis[v][w-a],v,w-a));
			}
		}
		int nw=min(2500,w+c[u]);
		if(dis[u][nw]>d+dd[u])
		{
			dis[u][nw]=d+dd[u];
			q.push(node(dis[u][nw],u,nw));
		}
	}
	for(int i=2;i<=n;i++)
	{
		LL res=INF;
		for(int j=0;j<=2500;j++) res=min(res,dis[i][j]);
		printf("%lld\n",res);
	}
	return 0;
}