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;

int n,m,s,d[505],sum;
int v[505],a[505],nxt[505],head[505];
long long dis[55][500005],ans[55],b[505],c[505];
bool vis[55][500005];

void spfa()
{
	queue<pair<int,int> > q;
	q.push(make_pair(1,s));
	while(!q.empty())
	{
		int u=q.front().first,x=q.front().second;
		q.pop(),vis[u][x]=0;
//		cout<<u<<' '<<x<<' '<<dis[u][x]<<endl;
		ans[u]=min(ans[u],dis[u][x]);
		for(int i=x+d[u],j=1;i<sum+d[u];i+=d[u],j++)
		{
			if(i>sum) i=sum;
			if(dis[u][i]>dis[u][x]+c[u]*j)
			{
				dis[u][i]=dis[u][x]+c[u]*j;
				if(!vis[u][i])
				{
					vis[u][i]=1,q.push(make_pair(u,i));
				}
			}
			else break;
			if(i==sum) break;
		}
		for(int i=head[u];i>0;i=nxt[i])
		{
			if(x<a[i]) continue;
			if(dis[v[i]][x-a[i]]>dis[u][x]+b[i])
			{
//				cout<<' '<<v[i]<<' '<<x-a[i]<<' '<<dis[u][x]+b[i]<<' '<<b[i]<<endl;
				dis[v[i]][x-a[i]]=dis[u][x]+b[i];
				if(!vis[v[i]][x-a[i]])
				{
					vis[v[i]][x-a[i]]=1;
					q.push(make_pair(v[i],x-a[i]));
				}
			}
		}
	}
}

int main()
{
	memset(ans,0x3f,sizeof(ans));
	scanf("%d%d%d",&n,&m,&s);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&v[i],&v[i+m]);
		nxt[i]=head[v[i+m]],head[v[i+m]]=i;
		nxt[i+m]=head[v[i]],head[v[i]]=i+m;
		scanf("%d%d",&a[i],&b[i]);
		a[i+m]=a[i],b[i+m]=b[i];
		sum+=a[i];
	}
	sum=sum*2,s=min(s,sum);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&d[i],&c[i]);
		d[i]=min(d[i],sum);
	}
	memset(dis,0x3f,sizeof(dis));
	dis[1][s]=0,vis[1][s]=1,spfa();
	for(int i=2;i<=n;i++) printf("%lld\n",ans[i]);
}