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 <cstdio>
#include <algorithm>
#define LL long long
#define INF 1e15
#define M 3000
using namespace std;
int n,m,s,head[55],tot,c[55],d[55];
LL dp[55][3010];
struct node{
	int to,next,a,b;
}e[105*2];
void add_edge(int u,int v,int a,int b){
	tot++,e[tot].to=v,e[tot].a=a,e[tot].b=b,e[tot].next=head[u],head[u]=tot;
}
int main(){
	int i,j,u,v,a,b,flag;
	scanf("%d%d%d",&n,&m,&s);
	for(i=1;i<=m;i++){
		scanf("%d%d%d%d",&u,&v,&a,&b);
		add_edge(u,v,a,b),add_edge(v,u,a,b);
	}
	for(i=1;i<=n;i++)scanf("%d%d",&c[i],&d[i]);
	for(i=1;i<=n;i++)
		for(j=0;j<=M;j++)dp[i][j]=INF;//注意j从0开始
	s=min(M,s);
	dp[1][s]=0;
	while(1){
		flag=0;
		for(u=1;u<=n;u++){//Bellman-Ford思想
			for(j=0;j<=M;j++){
				if(j>=c[u])dp[u][j]=min(dp[u][j],dp[u][j-c[u]]+d[u]);//01背包
				for(b=head[u];b;b=e[b].next){
					v=e[b].to;
					if(j>=e[b].a&&dp[v][j-e[b].a]>dp[u][j]+e[b].b){
						flag=1;
						dp[v][j-e[b].a]=dp[u][j]+e[b].b;
					}
				}
			}
			for(j=M-1;j>=0;j--)dp[u][j]=min(dp[u][j],dp[u][j+1]);//清空手上的钱，但不消耗时间
		}
		if(!flag)break;
	}
	for(i=2;i<=n;i++)printf("%lld\n",dp[i][0]);
	return 0;
}