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 tuple<int64_t, int64_t, int64_t> T;
const int64_t INF = (1LL << 62);
const int64_t MAXN = 55, MAXS = 2500;
int64_t N, M, S;
vector<T> to[MAXN];
vector<int64_t> C(MAXN), D(MAXN);

void dijkstra()
{
  priority_queue<T, vector<T>, greater<T>> PQ;
  vector<vector<int64_t>> dp(N, vector<int64_t>(MAXS + 1, INF));
  dp[0][S] = 0;
  PQ.emplace(0, 0, S);
  while (!PQ.empty())
  {
    int64_t time, v, s;
    tie(time, v, s) = PQ.top();
    PQ.pop();
    if (time > dp[v][s])
    {
      continue;
    }
    if (s + C.at(v) <= MAXS)
    {
      int64_t ns = s + C.at(v);
      int64_t ntime = time + D.at(v);
      if (ntime < dp[v][ns])
      {
        dp[v][ns] = ntime;
        PQ.emplace(ntime, v, ns);
      }
    }
    for (auto u : to[v])
    {
      int64_t nv, a, b;
      tie(nv, a, b) = u;
      if (s < a)
        continue;
      int64_t ns = s - a, ntime = time + b;
      if (ntime < dp[nv][ns])
      {
        dp[nv][ns] = ntime;
        PQ.emplace(ntime, nv, ns);
      }
    }
  }
  for (int v = 1; v < N; v++)
  {
    int64_t res = INF;
    for (int s = 0; s <= MAXS; s++)
    {
      res = min(res, dp[v][s]);
    }
    cout << res << endl;
  }
}

int main()
{
  cin >> N >> M >> S;
  if (S >= MAXS)
    S = MAXS;
  int64_t u, v, a, b;
  for (int i = 0; i < M; i++)
  {
    cin >> u >> v >> a >> b;
    u--, v--;
    to[u].emplace_back(v, a, b);
    to[v].emplace_back(u, a, b);
  }
  for (int i = 0; i < N; i++)
  {
    cin >> C.at(i) >> D.at(i);
  }
  dijkstra();
}