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;
typedef long long int ll;
typedef pair<ll,ll> pii;
#define fi first
#define se second
#define mod 1000000007
#define maxn 2500ll
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl "\n"
struct Cell
{
	ll v,cst,mt;
};
vector<Cell> adj[55];
ll dp[55][2505];
pii a[55];
void dijkstra(ll u,ll c)
{
	set<pair<ll,pii> >st;
	pii temp;
	temp.fi=u;temp.se=c;
	st.insert({0,temp});
	dp[u][c]=0;
	while(!st.empty())
	{
		pair<ll,pii> temp=*st.begin();
		st.erase(st.begin());
		ll mnt=temp.fi;
		u=(temp.se).fi;
		c=(temp.se).se;
		if(c<maxn)
		{
			ll val=min(maxn,c+a[u].fi);
			ll &res=dp[u][val];
			if(res>mnt+a[u].se)
			{
				st.erase({res,{u,val}});
				res=mnt+a[u].se;
				st.insert({res,{u,val}});
			}
		}
		for(auto it=adj[u].begin();it!=adj[u].end();it++)
		{
			Cell temp=*it;
			if(c>=temp.cst)
			{
				ll &res=dp[temp.v][c-temp.cst];
				if(res>mnt+temp.mt)
				{
					st.erase({res,{temp.v,c-temp.cst}});
					res=mnt+temp.mt;
					st.insert({res,{temp.v,c-temp.cst}});
				}
			}
		}
	}
}
int main()
{
	ios;
	ll i,m,s,n,u,j,k;
	cin>>n>>m>>s;
	s=min(s,maxn);
	for(i=1;i<=m;i++)
	{
		Cell temp;
		cin>>u>>temp.v>>temp.cst>>temp.mt;
		adj[u].push_back(temp);
		swap(temp.v,u);
		adj[u].push_back(temp);
	}
	for(i=1;i<=n;i++)
		cin>>a[i].fi>>a[i].se;
	for(i=1;i<=n;i++)
	{
		for(j=0;j<=maxn;j++)
			dp[i][j]=LLONG_MAX;
	}
	dijkstra(1,s);
	for(i=2;i<=n;i++)
	{
		ll min1=LLONG_MAX;
		for(j=0;j<=maxn;j++)
			min1=min(min1,dp[i][j]);
		cout<<min1<<endl;
	}
}
