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;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
using Int = long long;
const char newl = '\n';

//INSERT ABOVE HERE
const Int MAX = 55;
const Int INF = 1e18;
Int dp[MAX][MAX*MAX];

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);

  Int n,m,s_init;
  cin>>n>>m>>s_init;
  struct edge{
    Int to,a,b;
    edge(Int to,Int a,Int b):to(to),a(a),b(b){}
  };

  vector<vector<edge>> G(n);
  for(Int i=0;i<m;i++){
    Int u,v,a,b;
    cin>>u>>v>>a>>b;
    u--;v--;
    G[u].emplace_back(v,a,b);
    G[v].emplace_back(u,a,b);
  }

  vector<Int> cs(n),ds(n);
  for(Int i=0;i<n;i++) cin>>cs[i]>>ds[i];

  for(Int i=0;i<MAX;i++)
    for(Int j=0;j<MAX*MAX;j++)
      dp[i][j]=INF;

  using T = tuple<Int, Int, Int>;
  priority_queue<T, vector<T>, greater<T>> pq;

  auto push=[&](Int v,Int s,Int t){
    chmin(s,MAX*MAX-1);
    if(s<0) return;
    if(dp[v][s]<=t) return;
    dp[v][s]=t;
    pq.emplace(t,v,s);
  };
  push(0,s_init,0);

  while(!pq.empty()){
    auto [t,v,s]=pq.top();pq.pop();
    if(dp[v][s]<t) continue;
    // exchange
    {
      Int to=v;
      Int ns=s+cs[v];
      Int nt=t+ds[v];
      push(to,ns,nt);
    }
    for(auto [to, a, b]:G[v]){
      Int ns=s-a;
      Int nt=t+b;
      push(to,ns,nt);
    }
  }

  vector<Int> ans(n,INF);
  for(Int i=0;i<n;i++)
    for(Int j=0;j<MAX*MAX;j++)
      chmin(ans[i],dp[i][j]);

  for(Int i=1;i<n;i++) cout<<ans[i]<<newl;
  return 0;
}
