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 <iostream>
#include <queue>
#include <vector>
#include <tuple>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef tuple<ll, int, int> P; // time, node idx, money
vector<P> G[55];
int C[55];
ll D[55];
int arrived[55][2505];
ll RET[55][2505];
int main() {
  ios::sync_with_stdio(false);
  int N, M, U, V, A, S;
  ll B;
  cin >> N >> M >> S;
  for (int i=0; i<M; i++) {
    cin >> U >> V >> A >> B;
    G[U].push_back(make_tuple(B, V, A));
    G[V].push_back(make_tuple(B, U, A));
  }
  for (int i=1; i<=N; i++) {
    cin >> C[i] >> D[i];
  }
  priority_queue<P, vector<P>, greater<P> > q;
  int TH = 2500;
  q.push(make_tuple(0LL, 1, min(S,TH)));
  while(!q.empty()) {
    auto p = q.top(); q.pop();
    ll t = get<0>(p);
    int idx = get<1>(p);
    int mon = get<2>(p);
    if (arrived[idx][mon]) continue;
    arrived[idx][mon] = 1;
    RET[idx][mon] = t;
    if (mon < TH) {
      q.push(make_tuple(t + D[idx], idx, min(TH, mon + C[idx])));
    }
    for (int i=0; i<G[idx].size(); i++) {
      ll tt = get<0>(G[idx][i]);
      int ii = get<1>(G[idx][i]);
      ll mm = get<2>(G[idx][i]);
      if (mon < mm) continue;
      if (arrived[ii][mon-mm]) continue;
      q.push(make_tuple(t + tt, ii, mon - mm));
    }
  }
  for (int i=2; i<=N; i++) {
    ll ret = 1LL << 60;
    for (int j=0; j<=TH; j++) {
      ret = min(ret, RET[i][j]);
    }
    cout << ret << '\n';
  }
}