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;

int main() {
  ll n, m, s;
  cin >> n >> m >> s;
  vector<vector<tuple<int, ll, ll>>> g(n);
  for (int i = 0; i < m; i++) {
    ll u, v, a, b;
    cin >> u >> v >> a >> b;
    g[u - 1].push_back({v - 1, a, b});
    g[v - 1].push_back({u - 1, a, b});
  }
  vector<ll> c(n), d(n);
  for (int i = 0; i < n; i++) cin >> c[i] >> d[i];

  ll s_max = 50 * n;
  s = min(s_max, s);

  vector<vector<ll>> r(n, vector<ll>(s_max + 1, 1e18));
  r[0][s] = 0;

  queue<tuple<ll, ll, ll>> q;
  for (int i = 0;; i++) {
    q.push({0, min(s_max, s + i * c[0]), i * d[0]});
    if (s + i * c[i] >= s_max) break;
  }

  while (!q.empty()) {
    auto [i, s1, t1] = q.front();
    q.pop();
    // cerr << i << ' ' << s1 << ' ' << t1 << endl;
    for (auto [j, a, b] : g[i]) {
      // cerr << "  " << j << ' ' << a << ' ' << b << endl;
      for (int k = 0;; k++) {
        ll s2 = min(s_max, s1 + k * c[i] - a);
        ll t2 = t1 + k * d[i] + b;
        if (s2 >= 0 && t2 < r[j][s2]) {
          // cerr << "    " << j << ' ' << s2 << ' ' << t2 << endl;
          r[j][s2] = t2;
          q.push({j, s2, t2});
        }
        if (s2 >= s_max) break;
      }
    }
  }

  for (int i = 1; i < n; i++) cout << *min_element(r[i].begin(), r[i].end()) << endl;
}
