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> mp;
ll inf =  1e15;
typedef pair<ll,mp> mmp;// time , cost , now

int main(){
 ll n,m,s;
 cin>>n>>m>>s;
 

 vector<vector<mmp> > g(n);
 for(ll i=0;i<m;i++){
  ll u,v,a,b;
  cin>>u>>v>>a>>b;
  u--,v--;
  g[u].push_back( mmp( b, mp(a,v) ) );
  g[v].push_back( mmp( b, mp(a,u) ) );
 }

 vector<ll> c(n),d(n);
 for(ll i=0;i<n;i++){
  cin>>c[i]>>d[i];
 }
   //cout<<"OK"<<endl;
 ll M = (n+1)*50;
 s = min( s , M );
 vector<vector<ll> > dp( n , vector<ll>( M + 1 ,inf) );
 priority_queue<mmp,vector<mmp>,greater<mmp> > q;
 q.push( mmp( 0ll , mp( s , 0ll ) ) );
 dp[0][s] = 0;
 vector<ll> coin(n,0);

 while( !q.empty() ){
  mmp now = q.top();
  q.pop();
  ll nn = now.second.second;
  ll time = now.first;
  ll cost = now.second.first;
  //if( dp[nn][cost] != inf ) continue;
  //while( cost <= M ){
   do {
   for(ll i=0;i<g[nn].size();i++){
    ll next = g[nn][i].second.second;
    ll nec = min( M , cost - g[nn][i].second.first);
    ll ntime = time + g[nn][i].first;
    if( nec < 0 ) continue;
    if( dp[next][nec] <= ntime ) continue;
    //if( coin[next] > nec ) continue;
    dp[next][nec] = ntime;
    coin[next] = max( coin[i] , nec );
    q.push( mmp( ntime , mp( nec , next) ) );
   }
   cost+=c[nn];
   time+=d[nn];
  }while( cost <= M );
 // cout<<"OK"<<endl;
 }

 for(ll i=1;i<n;i++){
  ll ans = inf;
  for(ll j=0;j<M;j++) ans = min( ans , dp[i][j]);
  cout<<ans<<endl;
 }


    return 0;
}
