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 int N = 55;

int n, m, s;
int c[N], d[N];
vector<tuple<int, int, int>> edges[N];

long long dist[N][N*N];

int solve() {
  scanf("%d %d %d", &n, &m, &s);
  s = min(s, 2500);
  for (int i = 0; i < m; ++i) {
    int u, v, a, b;
    scanf("%d %d %d %d", &u, &v, &a, &b);
    --u, --v;

    edges[u].emplace_back(v, a, b);
    edges[v].emplace_back(u, a, b);
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &c[i], &d[i]);
  }

  memset(dist, -1, sizeof dist);

  priority_queue<tuple<long long, int, int>> pq;
  pq.push(make_tuple(0, 0, s));

  while (pq.size()) {
    long long dd;
    int u, s;
    tie(dd, u, s) = pq.top();
    pq.pop();

    if (dist[u][s] != -1) continue;
    dist[u][s] = -dd;

    {
      int ns = min(2500, s + c[u]);
      if (dist[u][ns] == -1) pq.push(make_tuple(dd - d[u], u, ns));
    }
    for (tuple<int, int, int> edge: edges[u]) {
      int v, a, b;
      tie(v, a, b) = edge;

      if (s - a < 0) continue;
      if (dist[v][s] != -1) continue;

      pq.push(make_tuple(dd - b, v, s-a));
    }
  }

  for (int i = 1; i < n; ++i) {
    long long mini = 1e18;
    for (int j = 0; j <= 2500; ++j) {
      if (dist[i][j] != -1) mini = min(mini, dist[i][j]);
    }

    printf("%lld\n", mini);
  }
  return 0;
}

int main() {
  int t = 1;
  // scanf("%d", &t);
  for (int tc = 0; tc < t; ++tc) {
    // printf("Case #%d: ", tc+1);
    solve();
  }
  return 0;
}
