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;
typedef long long ll;

const int nax = 50+10;
const ll M = 1e4;
vector<tuple<int,ll,ll>> node[nax];
ll dist[nax][M+10];

ll rate[nax], rtime[nax];

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  priority_queue<tuple<ll,ll,ll>> pq;
  pq.emplace(0, 0, min((ll)s, M));

  for (int i = 0; i < m; i++) {
    int a, b, cost, time;
    cin >> a >> b >> cost >> time;
    a--, b--;
    node[a].emplace_back(b, cost, time);
    node[b].emplace_back(a, cost, time);
  }
  for (int i = 0; i < n; i++) {
    cin >> rate[i] >> rtime[i];
  }

  fill_n((ll*)dist, nax*(M+10), 1e18);

  while (pq.size()) {
    auto [d, p, c] = pq.top();
    pq.pop();
    d *= -1;
    if (d > dist[p][c]) continue;
    for (auto [i, cost, time] : node[p]) {
      ll nd = d+time;
      ll nc = c-cost;
      if (nc >= 0 && nd < dist[i][nc]) {
	dist[i][nc] = nd;
	pq.emplace(-nd, i, nc);
      }
    }
    {
      ll nd = d+rtime[p];
      ll nc = min(M, c+rate[p]);
      if (nd < dist[p][nc]) {
	dist[p][nc] = nd;
	pq.emplace(-nd, p, nc);
      }
    }
  }

  for (int i = 1; i < n; i++) {
    ll ans = 1e18;
    for (int c = 0; c <= M; c++)
      ans = min(ans, dist[i][c]);
    cout << ans << '\n';
  }

}
