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;
using ll = long long;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  ll N; cin >> N;
  ll M; cin >> M;
  ll S; cin >> S;
  vector<ll> C(N), D(N);

  vector< pair<ll, pair<ll, ll>> > G[N]; // to, coin, time

  for (ll i = 0; i < M; i++) {
    ll u, v, a, b;
    cin >> u >> v >> a >> b;
    u--, v--;
    G[u].push_back(make_pair(v, make_pair(a, b)));
    G[v].push_back(make_pair(u, make_pair(a, b)));
  }
  for (ll i = 0; i < N; i++) {
    cin >> C[i] >> D[i];
  }

  S = min(S, (ll) 2500);
  ll INF = 1e18;
  vector<vector<ll>> dp(N, vector<ll>(2550, INF));
  queue< pair<ll, pair<ll, ll>> > q; // time, location, coin
  q.push(make_pair(0, make_pair(0, S)));
  while (!q.empty()) {
    pair<ll, pair<ll, ll>> c = q.front(); q.pop();
    ll mins = c.first, loc = c.second.first, coin = c.second.second;
    if (dp[loc][coin] <= mins) continue;
    dp[loc][coin] = mins;

    for (auto n: G[loc]) {
      ll new_mins = mins + n.second.second;
      ll new_loc = n.first;
      ll new_coin = coin - n.second.first;
      if (new_coin < 0) continue;
      q.push(make_pair(new_mins, make_pair(new_loc, new_coin)));
    }

    ll new_mins = mins + D[loc];
    ll new_coin = coin + C[loc];
    if (new_coin > 2500) new_coin = 2500;
    q.push(make_pair(new_mins, make_pair(loc, new_coin)));
  }

  for (ll i = 1; i < N; i++) {
    ll ans = INF;
    for (ll j = 0; j < 2550; j++) {
      ans = min(ans, dp[i][j]);
    }
    cout << ans << '\n';
  }
  return 0;
}