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;

struct Edge {
  long to, c, t;
};

const long inf = 1e18; 
const long mx = 5000;
long n, m, s;
vector<Edge> g[50];
long ca[50];
long ti[50];

vector<long> sub() {
  using PP = pair<long, long>;
  using TT = pair<long, PP>;
  
  priority_queue<TT, vector<TT>, greater<TT>> q;
  q.push(TT(0, PP(0, min(s, mx))));
  
  vector<vector<long>> dist(n, vector<long>(mx + 1, inf));
  while (!q.empty()) {
    auto tt = q.top(); q.pop();
    long d = tt.first;
    long from = tt.second.first;
    long mon = tt.second.second;
    if (dist[from][mon] != inf) continue;
    dist[from][mon] = d;
    for (auto e : g[from]) {
      long to = e.to;
      long c = e.c;
      long t = e.t;
      if (mon >= c) {
        q.push(TT(d + t, PP(to, mon - c)));
      }
    }
    q.push(TT(d + ti[from], PP(from, min(mon + ca[from], mx))));
  }
  
  vector<long> ret(n);
  for (int i = 0; i < n; ++i) {
    long mn = inf;
    for (int j = 0; j <= mx; ++j) {
      mn = min(mn, dist[i][j]);
    }
    ret[i] = mn;
  }
  return ret;
}

int main() {
  cin >> n >> m >> s;
  for (int j = 0; j < m; ++j) {
     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});
  }
  for (int i = 0; i < n; ++i) {
    cin >> ca[i] >> ti[i]; 
  }
  
  auto v = sub();
  for (int i = 1; i < n; ++i) {
    cout << v[i] << endl;
  }
}