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;
using ll = long long;

#define REP(i, n) for(int (i) = 0; (i) < (n); ++(i))
#define REPR(i, n) for(int (i) = (n); (i) >= 0; --(i))
#define FOR(i, n, m) for(int (i) = (n); (i) < (m); ++(i))

// constexpr int INF = 1e9;
constexpr ll INF = 1LL<<61;
constexpr ll mod = 1e9+7;
int maxs = 2500;

vector<ll> dijkstra(vector<vector<pair<int, ll>>> g, int s){
  int n = g.size();
  vector<ll> dist(n, INF);
  dist[s] = 0;
  priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
  pq.push({0, s});
  while(!pq.empty()){
    auto[d, v] = pq.top(); pq.pop();
    if(d > dist[v]) continue;
    for(auto e : g[v]){
      auto[nv, c] = e;
      if(dist[nv] > dist[v]+c){
        dist[nv] = dist[v]+c;
        pq.push({dist[nv], nv});
      }
    }
  }
  return dist;
}
int main(){
  int N, M, S;
  cin >> N >> M >> S;
  S = min(S, 2499);
  vector<vector<pair<int, ll>>> g(N*2500);
  REP(i, M){
    int U, V, A, B;
    cin >> U >> V >> A >> B;
    U--, V--;
    FOR(j, A, maxs){
      g[U*maxs+j].push_back({V*maxs+j-A, B});
      g[V*maxs+j].push_back({U*maxs+j-A, B});
    }
  }
  REP(i, N){
    int C, D;
    cin >> C >> D;
    REP(j, maxs-C){
      g[i*maxs+j].push_back({i*maxs+j+C, D});
    }
  }
  vector<ll> dist = dijkstra(g, S);
  vector<ll> ans(N, INF);
  REP(i, N){
    REP(j, 2500){
      ans[i] = min(ans[i], dist[i*maxs+j]);
    }
  }
  FOR(i, 1, N){
    cout << ans[i] << endl;
  }
  return 0;
}
