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 vec = vector<pair<int, int64_t>>;
using graph = vector<vec>;
constexpr int64_t inf = 1e18;

int main() {
  int n, m, s;
  cin >> n >> m >> s;
  vector<int> u(m), v(m), a(m), b(m), c(n), d(n);
  for (int i = 0; i < m; i++) {
    cin >> u.at(i) >> v.at(i) >> a.at(i) >> b.at(i);
    u.at(i)--;
    v.at(i)--;
  }
  for (int i = 0; i < n; i++) {
    cin >> c.at(i) >> d.at(i);
  }
  graph to(50 * n * n);
  for (int i = 0; i < m; i++) {
    for (int j = a.at(i); j < 50 * n; j++) {
      to.at(50 * n * u.at(i) + j).emplace_back(50 * n * v.at(i) + j - a.at(i), b.at(i));
      to.at(50 * n * v.at(i) + j).emplace_back(50 * n * u.at(i) + j - a.at(i), b.at(i));
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 50 * n; j++) {
      int next = min(50 * n - 1, j + c.at(i));
      to.at(50 * n * i + j).emplace_back(50 * n * i + next, d.at(i));
    }
  }
  vector<int64_t> cost(50 * n * n, inf);
  priority_queue<pair<int64_t, int>> q;
  q.emplace(0, min(s, 50 * n - 1));
  while (!q.empty()) {
    auto x = q.top();
    q.pop();
    if (cost.at(x.second) < inf) continue;
    cost.at(x.second) = -x.first;
    for (auto next : to.at(x.second)) {
      if (cost.at(next.first) < inf) continue;
      q.emplace(x.first - next.second, next.first);
    }
  }
  for (int i = 1; i < n; i++) {
    int64_t ans = inf;
    for (int j = 0; j < 50 * n; j++) {
      ans = min(ans, cost.at(50 * n * i + j));
    }
    cout << ans << endl;
  }
}
