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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

typedef long long ll;

int n, m, s;
ll a[51][51], b[51][51];
ll c[51];
ll d[51];

int maxa = 0;

struct Data {
  ll t, s, p;
  bool operator<(const Data& d)const{return t > d.t;}
};

ll dijkstra(int g) {
  priority_queue< Data > Q;
  bool visited[51][2501] = {false};
  Q.push({0, min(s, maxa * (n-1)), 1});

  while(!Q.empty()) {
    Data now = Q.top();
    Q.pop();

    if (now.p == g) {
      return now.t;
    }
    if (visited[now.p][now.s]) continue;
    visited[now.p][now.s] = true;

    if (now.s + c[now.p] <= maxa * (n-1)) {
      Q.push({now.t + d[now.p], now.s + c[now.p], now.p});
    }

    for (int j = 1; j <= n; j++){
      if (a[now.p][j] == 0) continue;

      int ns = now.s - a[now.p][j];
      if (ns < 0) continue;
      Q.push({now.t + b[now.p][j], ns, j});
    }
  }
}

int main() {
  cin >> n >> m >> s;

  for (int i = 0; i < m; i++) {
    int u, v, aa, bb;
    cin >> u >> v >> aa >> bb;
    a[u][v] = a[v][u] = aa;
    b[u][v] = b[v][u] = bb;
    maxa = max(maxa, aa);
  }
  for (int i = 1; i <= n; i++){
    cin >> c[i] >> d[i];
  }

  for (int i = 2; i <= n; i++) {
    cout << dijkstra(i) << endl;
  }

  return 0;
}
