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;

int n, m;
long long s;
unsigned long long ans[55];
unsigned long long dp[55][7005];
pair<int, int> refill[200005];
vector<pair<int, pair<long, long> > > g[55];

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

  cin >> n >> m >> s;
  for(int i = 0; i < m; i++) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    g[u-1].push_back(make_pair(v-1, make_pair(a, b)));
    g[v-1].push_back(make_pair(u-1, make_pair(a, b)));
  }

  for(int i = 0; i < n; i++) {
    int c, d;
    cin >> c >> d;
    refill[i] = make_pair(c, d);
  }

  for(int i = 0; i < n; i++) for(int j = 0; j < 7005; j++) dp[i][j] = 10e12;

  dp[0][(min(s, 7004ll))] = 0;
  for(int i = 0; i < 700; i++) {
    for(int j = 0; j < n; j++) 
      for(int k = 0; k < 7005-refill[j].first; k++) 
        dp[j][k+refill[j].first] = min(dp[j][k+refill[j].first], dp[j][k]+refill[j].second);

    for(int j = 0; j < n; j++) 
      for(int k = 0; k < 7005; k++) 
        for(auto& a : g[j]) 
          if(k >= a.second.first && dp[j][k] < 10e12) 
            dp[a.first][k-a.second.first] = min(dp[a.first][k-a.second.first], dp[j][k]+a.second.second);
  }

  for(int i = 0; i < n; i++) ans[i] = dp[i][0];
  for(int i = 0; i < n; i++) for(int j = 0; j < 7005; j++) ans[i] = min(ans[i], dp[i][j]);
  for(int i = 1; i < n; i++) cout << ans[i] << endl;
}