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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;
using info = pair<ll,P>;

vector<pair<int,P>> g[50];
int main() {
  int n,m,s;
  cin >> n >> m >> s;
  rep(i,m){
    int u,v,a,b;
    cin >> u >> v >> a >> b;
    --u; --v;
    g[u].emplace_back(v,P(a,b));
    g[v].emplace_back(u,P(a,b));
  }
  vector<int> c(n), d(n);
  rep(i,n) cin >> c[i] >> d[i];

  vector<vector<ll>> dp(n,vector<ll>(2505, 1e18L));
  priority_queue<info,vector<info>,greater<info>> q;
  auto push = [&](int v, int s, ll t){
    if(s<0) return;
    if(dp[v][s]<=t) return;
    dp[v][s] = t;
    q.emplace(t,P(v,s));
  };
  push(0,min(s,2500),0);
  while(!q.empty()){
    info info = q.top(); q.pop();
    int u = info.second.first;
    int s = info.second.second;
    ll t = info.first;

    if(dp[u][s]!=t) continue;
    push(u,min(s+c[u],2500),t+d[u]);
    for(pair<int,P> e:g[u]){
      int v = e.first;
      P p = e.second;
      push(v,s-p.first,t+p.second);
    }
  }
  
  rep(i,n-1){
    ll ans = 1e18L;
    for(ll t:dp[i+1]) ans = min(ans,t);
    cout << ans << endl;
  }
}
