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 N 55
#define S 5005
#define ll long long
#define set(a,v) memset(a,v,sizeof(a))
int n,m,s,lim=5e3,c[N],d[N];bool vs[N];ll dp[N][S];
struct Edge {int v,a,b;};vector<Edge> e[N];queue<int> q;
void addE(int u,int v,int a,int b)
{
	e[u].push_back((Edge) {v,a,b});
	e[v].push_back((Edge) {u,a,b});
}
void SPFA()
{
	set(dp,0x3f);dp[1][s]=0;vs[1]=1;q.push(1);
	while(!q.empty())
	{
		int u=q.front();q.pop();vs[u]=0;
		for(int i=c[u];i<=lim;++i)
			dp[u][i]=min(dp[u][i],dp[u][i-c[u]]+d[u]);
		for(int i=0,v,a,b,fl;i<e[u].size();++i)
		{
			v=e[u][i].v;a=e[u][i].a;b=e[u][i].b;fl=0;
			for(int j=a;j<=lim;++j) if(dp[u][j]+b<dp[v][j-a])
				dp[v][j-a]=dp[u][j]+b,fl=1;
			if(fl && !vs[v]) vs[v]=1,q.push(v);
		}
	}
}
int main()
{
	scanf("%d %d %d",&n,&m,&s);s=min(s,lim);
	for(int i=1,u,v,a,b;i<=m;++i)
		scanf("%d %d %d %d",&u,&v,&a,&b),addE(u,v,a,b);
	for(int i=1;i<=n;++i) scanf("%d %d",&c[i],&d[i]);SPFA();
	for(int i=2;i<=n;++i)
	{
		ll res=1e18;
		for(int j=0;j<=lim;++j) res=min(res,dp[i][j]);
		printf("%lld\n",res);
	}return 0;
}