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;
using ll=long long;
struct E{ll to,co,ti;};
struct P{ll ti,id,re;};
bool operator <(const P &a,const P &b){return a.ti>b.ti;}
ll dp[50][5001],c[50],d[50],ans[50],n,m,s;
vector<E> edge[50];
priority_queue<P> que;

signed main(){
  cin>>n>>m>>s;
  while(m--){
    ll u,v,a,b;cin>>u>>v>>a>>b;u--;v--;
    edge[u].push_back({v,a,b});
    edge[v].push_back({u,a,b});
  }
  for(ll i=0;i<n;i++)cin>>c[i]>>d[i];
  que.push({1,0,min(5000LL,s)});
  while(que.size()){
    P p=que.top();que.pop();
    if(dp[p.id][p.re])continue;
    dp[p.id][p.re]=p.ti;
    if(!ans[p.id])ans[p.id]=p.ti;
    for(E q:edge[p.id])
      if(p.re>=q.co&&!dp[q.to][p.re-q.co])
        que.push({p.ti+q.ti,q.to,p.re-q.co});
    if(p.re+c[p.id]<=5000)que.push({p.ti+d[p.id],p.id,p.re+c[p.id]});
  }
  for(ll i=1;i<n;i++)cout<<--ans[i]<<endl;
}
