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 <iostream>
#include <cmath>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <set>
#include <algorithm>
#include <iomanip>
#include <string.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()

typedef long long lint;

using namespace std;
lint L=5001;
struct djk{
  djk(int n):n(n),c(n),d(n,inf){
  }
  void add(int s,int t,lint v){
    c[s].push_back(make_pair(v,t));
  }
  void solve(int start){
    q.push(make_pair(0,start));
    d[start]=0;
    while(!q.empty()){
      auto p=q.top();
      q.pop();
      lint pv=p.first;
      int pt=p.second;
      if(d[pt]>pv)continue;
      for(auto e:c[pt]){
        lint ev=e.first;
        int et=e.second;
        if(d[et]>d[pt]+ev){
          d[et]=d[pt]+ev;
          q.push(make_pair(d[et],et));
        }
      }
    }
  }
  const lint inf =1e15;
  int n;
  using edge=pair<lint,int>;
  vector<vector<edge> > c;
  vector<lint> d;
  priority_queue<edge,vector<edge>,greater<edge> > q;
};

int main(){
  lint N,M,S;
  cin>>N>>M>>S;
  djk djk(N*L);
  REP(i,M){
    lint U,V,A,B;
    cin>>U>>V>>A>>B;
    U--;V--;
    REP(l,L)if(l>=A){
      djk.add(U*L+l,V*L+l-A,B);
      djk.add(V*L+l,U*L+l-A,B);
    }
  }
  REP(i,N){
    lint C,D;
    cin>>C>>D;
    REP(l,L){
      djk.add(i*L+l,i*L+min(L-1,l+C),D);
    }
  }
  djk.solve(min(S,L-1));
  FOR(i,1,N){
    lint ans=djk.d[i*L];
    REP(l,L)ans=min(ans,djk.d[i*L+l]);
    cout<<ans<<endl;
  }
  return 0;
}
