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;

const int MAX_V=2501*50;
const int64_t INF=1e15;

struct edge{
  int to;
  int64_t cost;
};

typedef pair<int64_t,int> P;

vector<edge> G[MAX_V];
int64_t d[MAX_V];
int V;

void dijkstra(int s){

  priority_queue<P,vector<P>,greater<P>> que;
  fill(d,d+V,INF);
  d[s]=0;
  que.push(P(0,s));

  while(!que.empty()){
    P p=que.top();
    que.pop();
    int v=p.second;

    if(d[v]<p.first) continue;
    for(int i=0;i<G[v].size();++i){
      edge e=G[v][i];
      if(d[e.to]>d[v]+e.cost){
        d[e.to]=d[v]+e.cost;
        que.push(P(d[e.to],e.to));
      }
    }
  }

}

int main(){
  int64_t N,M,S;
  cin>>N>>M>>S;

  V=2501*N;

  for(int i=0;i<M;++i){
    int64_t u,v,a,b;
    cin>>u>>v>>a>>b;
    --u,--v;
    for(int j=a;j<=2500;++j){
      int x=2501*u+j;
      int y=2501*v+(j-a);
      G[x].push_back(edge{y,b});
    }
    for(int j=a;j<=2500;++j){
      int x=2501*v+j;
      int y=2501*u+(j-a);
      G[x].push_back(edge{y,b});
    }
  }

  for(int i=0;i<N;++i){
    int64_t c,d;
    cin>>c>>d;
    for(int j=0;j<=2500;++j){
      int x=2501*i+j;
      int y;
      if(j<=2500-c) y=2501*i+(j+c);
      else y=2501*i+2500;
      G[x].push_back(edge{y,d});
    }
  }
  
  if(S>=2500) dijkstra(2500);
  else dijkstra((int)S);

  for(int i=1;i<N;++i){
    int64_t ans=1e15;
    for(int j=0;j<=2500;++j){
      int x=2501*i+j;
      ans=min(ans,d[x]);
    }
    cout<<ans<<endl;
  }

}