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 pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second
 
const int mx=100010;
const ll mod=1e9+7;

typedef pair<ll,pair<int,int>> pli;

int main(){
  int n,m,s;
  cin >> n >> m >> s;
  struct info{
    int to; int cost; int time;
  };
  
  vector<int> c(n),d(n);
  vector<info> v[n];
  rep(i,m){
    int a,b,p,q;
    cin >> p >> q >> a >> b;
    p--; q--;
    v[p].push_back({q,a,b});
    v[q].push_back({p,a,b});
  }
  rep(i,n){
    cin >> c[i] >> d[i];
  }

  vector<vector<ll>> dist(n, vector<ll>(2510,longinf)); //city silvercoin
  priority_queue<pli,vector<pli>,greater<pli>> q;
  dist[0][min(2500,s)]=0;
  q.push({0LL,{0,min(2500,s)}});  //dist city money

  while(q.size()){
    pli p=q.top(); q.pop();
    int now = p.S.F;
    int left = p.S.S;
    if(dist[now][left]!=p.F) continue;
    auto go = [&](int nxt, int nleft, int cost){
      if(dist[nxt][nleft]>p.F+cost) q.push({dist[nxt][nleft]=p.F+cost,{nxt,nleft}});
    };
    go(now, min(2500,left+c[now]), d[now]);
    for(auto it:v[now]){
      if(it.cost > left) continue;
      go(it.to, left-it.cost, it.time);
    }
  }

  REP(i,1,n){
    ll ans = longinf;
    rep(j,2501) ans = min(ans, dist[i][j]);
    cout << ans << endl;
  }
  return 0;
}