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;
#define rep(i, n) for(int i=0; i<(n); i++)
#define PQ priority_queue<vector<ll>,vector<vector<ll>>,greater<vector<ll>>>
using ll = long long;
const ll INF=1e18;
const ll maxS = 3000;
 
map<int, ll> dij(int f,ll s, ll n, vector<vector<vector<ll>>>& edge){
  map<int, ll> mintime;
  vector<vector<ll>> mincost(maxS+5, vector<ll>(n, INF));
  s = min(s,maxS);
  mincost[s][f] = 0;
  
  PQ pq;
  pq.push({mincost[s][f], s, f});
  
  while(!pq.empty()){
    vector<ll> now=pq.top(); pq.pop();
	if(!mintime.count(now[2])) mintime[now[2]] = now[0];
    if((ll)mintime.size()==n) return mintime;
    
    vector<vector<ll>>& v = edge[now[2]];
    ll l = v.size();
    
    rep(i, l){
      ll nextS=min(now[1]+v[i][1], maxS);
      if(nextS < 0) continue;
 
      if(mincost[nextS][v[i][0]] <= now[0] + v[i][2]) continue;
      mincost[nextS][v[i][0]] = now[0] + v[i][2];
	  pq.push({mincost[nextS][v[i][0]], nextS, v[i][0]});
    }
  }
}
 
int main(){
  int n, m; cin >> n >> m;
  ll s; cin >> s;
 
  vector<vector<vector<ll>>> edge(n);  
  rep(i, m){
    int u, v; cin >> u >> v; 
    ll a, b;  cin >> a >> b;
    u--, v--;
    edge[u].push_back({v, -a, b});
    edge[v].push_back({u, -a, b});
    }
  rep(i, n){
    ll c, d; cin >> c >> d;
    edge[i].push_back({i, c, d});
  }
 
map<int, ll> mintime =  dij(0, s, n, edge);
  for(auto p : mintime){
    if(p.first == 0) continue;
    cout << p.second << endl;
  }
  return 0;
}