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 <iostream>
#include <queue>
using namespace std;
struct edge
{
	int v,n;
	long long w;
}e[2000005];
int head[500005],cnt;
long long dis[500005],ans[5005];
int vis[500005];
inline void add(int u,int v,long long w)
{
	e[++cnt]={v,head[u],w};
	head[u]=cnt;
}
priority_queue <pair<long long,int>,vector <pair<long long,int> >,greater <pair<long long,int> > > q;
inline int id(int x,int y)
{
	return x*5001+y;
}
int main(int argc, char** argv) {
	int n,m,st;
	cin >> n >> m >> st;
	for(int i=1;i<=m;i++)
	{
		int u,v,a,b;
		cin >> u >> v >> a >> b;
		for(int j=a;j<=5000;j++)
			add(id(u,j),id(v,j-a),b),add(id(v,j),id(u,j-a),b);
	}
	for(int i=1;i<=n;i++)
	{
		int c,d;
		cin >> c >> d;
		for(int j=0;j<=5000;j++)
			add(id(i,j),id(i,min(5000,j+c)),d);
	}
	int s=id(1,min(st,5000));
	q.push({0,s});
	for(int i=1;i<=500000;i++)
		dis[i]=1e18;
	dis[s]=0;
	while(!q.empty())
	{
		int x=q.top().second;
		q.pop();
		if(vis[x]) continue;
		vis[x]=1;
		for(int i=head[x];i;i=e[i].n)
		{
			if(dis[e[i].v]>dis[x]+e[i].w)
			{
				dis[e[i].v]=dis[x]+e[i].w;
				q.push({dis[e[i].v],e[i].v});
			}
		}
	}
	for(int i=2;i<=n;i++)
	{
		ans[i]=1e18;
		for(int j=0;j<=5000;j++)
			ans[i]=min(ans[i],dis[id(i,j)]);
		cout << ans[i] << "\n";
	}
//	for(int i=1;i<=n;i++)
//		cout << dis[i] << " ";
	return 0;
}