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;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repr(i, n) for (int i = (n) - 1; i >= 0; i--)
#define range(a) a.begin(), a.end()

ll dp[50][5050];

bool chmin(ll &x, ll y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}

struct edge {
  int v;
  ll a, b;
};

template<class T>
using heap = priority_queue<T, vector<T>, greater<T>>;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  int N, M, S; cin >> N >> M >> S;
  vector<vector<ll>> dp(N, vector<ll>(5050, LLONG_MAX));
  vector<vector<edge>> G(N);
  rep(_, M) {
    int u, v, a, b; cin >> u >> v >> a >> b; u--; v--;
    G[u].push_back((edge){v, a, b});
    G[v].push_back((edge){u, a, b});
  }
  vector<ll> C(N), D(N);
  rep(i, N) cin >> C[i] >> D[i];
  heap<tuple<ll, int, int>> hp;
  auto ins = [&](int u, int x, ll d) {
    if (chmin(dp[u][x], d)) {
      hp.emplace(d, u, x);
    }
  };
  ins(0, min(S, 5040), 0);
  while (not hp.empty()) {
    ll d;
    int u, x;
    tie(d, u, x) = hp.top(); hp.pop();
    if (dp[u][x] < d) continue;
    ins(u, min(5040LL, x + C[u]), dp[u][x] + D[u]);
    for (edge e : G[u]) {
      if (x >= e.a) {
        ins(e.v, x - e.a, dp[u][x] + e.b);
      }
    }
  }
  for (int i = 1; i < N; i++) {
    ll ans = LLONG_MAX;
    rep(j, 5050) {
      chmin(ans, dp[i][j]);
    }
    cout << ans << '\n';
  }
}