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 long long ll;
typedef pair<int, int> ii;
typedef tuple<ll, ll, ll> iii;

const int N = 51;
const int M = 2460;
int n, m, s, u, v, a, b;
int C[N], D[N];
vector<iii> e[N];
ll dp[N][M];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(dp, 0x3f, sizeof dp);

  cin >> n >> m >> s;
  s = min(s, 2450);
  for (int i = 0; i < m; ++i) {
    cin >> u >> v >> a >> b;
    e[u].push_back(iii(v, a, b)); // to, money, time
    e[v].push_back(iii(u, a, b));
  }

  for (int i = 1; i <= n; ++i) {
    cin >> C[i] >> D[i];
  }

  dp[1][s] = 0;
  priority_queue<iii, vector<iii>, greater<iii>> pq;
  pq.push(iii(0, 1, s));
  while (pq.size()) {
    iii cur = pq.top(); pq.pop();
    // time, point, money
    ll ct = get<0>(cur), u = get<1>(cur), cm = get<2>(cur);
    for (auto [v, m, t] : e[u]) {
      if (cm >= m && dp[v][cm - m] > ct + t) {
        dp[v][cm - m] = ct + t;
        pq.push(iii(ct + t, v, cm - m));
      }
    }
    if (dp[u][min(2450LL, cm + C[u])] > ct + D[u]) {
      dp[u][min(2450LL, cm + C[u])] = ct + D[u];
      pq.push(iii(ct + D[u], u, min(2450LL, cm + C[u])));
    }
  }

  for (int i = 2; i <= n; ++i) {
    ll ans = 0x3f3f3f3f3f3f3f3f;
    for (int j = 0; j <= 2450; ++j) {
      ans = min(ans, dp[i][j]);
    }
    cout << ans << endl;
  }

  return 0;
}
