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 ll;
typedef pair<int,int> P;
#define rep(i,m,n) for(int i=m;i<n;i++)
ll mod=1e9+7;

struct edge{
  int to;
  int cost;
  int time;
};

struct state{
  ll time;
  int to;
  int cost;
  bool operator < (const state &s) const{
    return time>s.time;
  }
};

bool comp(const state& s1,const state& s2){
  return s1.time<s2.time;
}

int main(){
  int n,m,s;
  cin>>n>>m>>s;
  s=min(s,50*n);
  vector<edge> path[n];
  rep(i,0,m){
    int u,v,a,b;
    cin>>u>>v>>a>>b;
    path[u-1].push_back(edge{v-1,a,b});
    path[v-1].push_back(edge{u-1,a,b});
  }
  int c[n],d[n];
  rep(i,0,n) cin>>c[i]>>d[i];
  priority_queue<state> q;
  q.push(state{0,0,s});
  ll dp[n][50*n+1];
  rep(i,0,n){
    rep(j,0,50*n+1){
      dp[i][j]=1e15;
    }
  }
  dp[0][s]=0;
  while(q.size()){
    state f=q.top();
    q.pop();
    if(dp[f.to][f.cost]<f.time) continue;
    if(dp[f.to][min(f.cost+c[f.to],50*n)]>f.time+d[f.to]){
      dp[f.to][min(f.cost+c[f.to],50*n)]=f.time+d[f.to];
      q.push(state{f.time+d[f.to],f.to,min(f.cost+c[f.to],50*n)});
    }
    for(edge e:path[f.to]){
      if(f.cost<e.cost) continue;
      if(dp[e.to][f.cost-e.cost]>f.time+e.time){
        dp[e.to][f.cost-e.cost]=f.time+e.time;
        q.push(state{dp[e.to][f.cost-e.cost],e.to,f.cost-e.cost});
      }
    }
  }
  rep(i,1,n){
    ll ans=1e15;
    rep(j,0,50*n+1) {
      ans=min(ans,dp[i][j]);
    }
    cout<<ans<<endl;
  }
}