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;

const int64_t INF = INT64_C(1) << 60;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, M, S;
  cin >> N >> M >> S;
  vector<vector<pair<int, int>>> G(N);
  vector<int> A(M), B(M);
  for (int i = 0; i < M; i++) {
    int u, v;
    cin >> u >> v >> A[i] >> B[i];
    u--, v--;
    G[u].emplace_back(make_pair(v, i));
    G[v].emplace_back(make_pair(u, i));
  }
  vector<int> C(N), D(N);
  for (int i = 0; i < N; i++) {
    cin >> C[i] >> D[i];
  }
  int mx = 3000;
  vector<vector<int64_t>> T(N, vector<int64_t>(mx + 1, INF));
  priority_queue<pair<int64_t, pair<int, int>>,
                 vector<pair<int64_t, pair<int, int>>>,
                 greater<pair<int64_t, pair<int, int>>>>
      que;
  que.push(make_pair(0, make_pair(0, min(mx, S))));
  while (!que.empty()) {
    auto q = que.top();
    que.pop();
    int64_t t = q.first;
    int u = q.second.first, c = q.second.second;
    if (T[u][c] != INF) continue;
    T[u][c] = t;
    que.push(make_pair(t + D[u], make_pair(u, min(mx, c + C[u]))));
    for (auto g : G[u]) {
      int v = g.first, a = A[g.second], b = B[g.second];
      if (a > c) continue;
      que.push(make_pair(t + b, make_pair(v, c - a)));
    }
  }
  for (int i = 1; i < N; i++) {
    cout << *min_element(T[i].begin(), T[i].end()) << '\n';
  }
  return 0;
}
