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;
using ll=long long;
#define MAX 300003
#define MOD 1000000007
#define INF 1000000000000000
using ppp=pair<ll,pair<int,int>>;

int main(){
  ll N,M,S;
  cin>>N>>M>>S;
  vector<vector<pair<int,int>>> G(N);
  vector<ll> A(M),B(M);
  for(int i=0;i<M;i++){
    int u,v;
    cin>>u>>v>>A.at(i)>>B.at(i);
    u--;
    v--;
    G.at(u).push_back(make_pair(v,i));
    G.at(v).push_back(make_pair(u,i));
  }
  vector<ll> C(N),D(N);
  for(int i=0;i<N;i++){
    cin>>C.at(i)>>D.at(i);
  }

  vector<vector<ll>> time(N,vector<ll>(2501,INF));
  if(S>2500){
    S=2500;
  }
  time.at(0).at(S)=0;
  priority_queue<ppp,vector<ppp>,greater<ppp>> pq;
  pq.push(make_pair(0,make_pair(0,S)));
  while(!pq.empty()){
    int v=pq.top().second.first;
    int money=pq.top().second.second;
    pq.pop();
    int nm=(int)min<ll>(2500,money+C.at(v));
    if(time.at(v).at(nm)>time.at(v).at(money)+D.at(v)){
      time.at(v).at(nm)=time.at(v).at(money)+D.at(v);
      pq.push(make_pair(time.at(v).at(nm),make_pair(v,nm)));
    }
    for(auto p:G.at(v)){
      int nv=p.first;
      int e=p.second;
      if(money<A.at(e)){
        continue;
      }
      if(time.at(nv).at(money-A.at(e))>time.at(v).at(money)+B.at(e)){
        time.at(nv).at(money-A.at(e))=time.at(v).at(money)+B.at(e);
        pq.push(make_pair(time.at(nv).at(money-A.at(e)),make_pair(nv,money-A.at(e))));
      }
    }
  }

  for(int i=1;i<N;i++){
    ll ans=INF;
    for(int j=0;j<=2500;j++){
      ans=min<ll>(ans,time.at(i).at(j));
    }
    cout<<ans<<endl;
  }
}
