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 main() {
  int64_t V,E,s,A=0;
  cin>>V>>E>>s;
  vector<vector<tuple<int64_t,int64_t,int64_t>>> G(V);
  for(int i=0;i<E;i++){
    int64_t a,b,c;
    int64_t d;
    cin>>a>>b>>c>>d;
    a--,b--;
    G.at(a).push_back(make_tuple(b,c,d));
    G.at(b).push_back(make_tuple(a,c,d));
    A=max(A,c);
  }
  vector<vector<int64_t>> r(V,vector<int64_t>(A*V+1,-1));
  vector<int64_t> z(V,-1);
  vector<pair<int64_t,int64_t>> T(V);
  for(int i=0;i<V;i++){
    int64_t c;
    int64_t d;
    cin>>c>>d;
    T.at(i)=make_pair(c,d);
  }
  if(s>A*V){
    s=A*V;
  }
  priority_queue<tuple<int64_t,int64_t,int64_t>,vector<tuple<int64_t,int64_t,int64_t>>,greater<tuple<int64_t,int64_t,int64_t>>> pq;
  pq.push(make_tuple(0,0,s));
  r.at(0).at(s)=0;
  while(!pq.empty()){
    int64_t t,a,m;
    tie(t,a,m)=pq.top();
    pq.pop();
    if(r.at(a).at(m)<t){
      continue;
    }
    if(z.at(a)==-1){
      z.at(a)=t;
    }
    for(tuple<int64_t,int64_t,int64_t> x:G.at(a)){
      int64_t b,g,k;
      tie(b,g,k)=x;
      if(m>=g){
        if(r.at(b).at(m-g)==-1||r.at(b).at(m-g)>t+k){
          r.at(b).at(m-g)=t+k;
          pq.push(make_tuple(t+k,b,m-g));
        }
      }
    }
    if(r.at(a).at(min(A*V,m+T.at(a).first))>t+T.at(a).second||r.at(a).at(min(A*V,m+T.at(a).first))==-1){
    r.at(a).at(min(A*V,m+T.at(a).first))=t+T.at(a).second;
    pq.push(make_tuple(t+T.at(a).second,a,min(A*V,m+T.at(a).first)));
    }
  }
  for(int i=1;i<V;i++){
    cout<<z.at(i)<<endl;
  }
  

}


