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 int long long
#define piii pair<int,pair<int,int> >
int n,m,s,c[55],d[55],dp[55][2525];
vector<piii>v[55];
signed main()
{
	cin>>n>>m>>s;
	for(int i=0;i<m;i++)
	{
		int a,b,c,d;
		cin>>a>>b>>c>>d;
		a--;
		b--;
		v[a].push_back(make_pair(b,make_pair(c,d)));
		v[b].push_back(make_pair(a,make_pair(c,d)));
	}
	for(int i=0;i<n;i++)
	{
		cin>>c[i]>>d[i];
	}
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<2510;j++)
		{
			dp[i][j]=1e18;
		}
	}
	queue<piii>q;
	q.push(make_pair(0ll,make_pair(min(n*50-1,s),0ll)));
	while(!q.empty())
	{
		piii nw=q.front();
		q.pop();
		if(nw.second.first>=0 && nw.second.second>=0 && nw.second.second<dp[nw.first][min(n*50-1,nw.second.first)])
		{
			dp[nw.first][min(n*50-1,nw.second.first)]=nw.second.second;
			for(int i=0;i<v[nw.first].size();i++)
			{
				piii pp=v[nw.first][i];
				q.push(make_pair(pp.first,make_pair(nw.second.first-pp.second.first,nw.second.second+pp.second.second)));
			}
			q.push(make_pair(nw.first,make_pair(nw.second.first+c[nw.first],nw.second.second+d[nw.first])));
		}
	}
	for(int i=1;i<n;i++)
	{
		int ans=1e18;
		for(int j=0;j<n*50;j++)
		{
			ans=min(ans,dp[i][j]);
		}
		cout<<ans<<endl;
	}
	return 0;
}