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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

typedef pair<pll,ll> ppll;

int main() {
  ll n,m,s;
  cin >> n >> m >> s;
  vector<vector<ppll>> G(n);
  for(ll i=0;i<m;i++) {
    ll u,v,a,b;
    cin >> u >> v >> a >> b;
    u--,v--;
    G[u].push_back({{v,a},b});
    G[v].push_back({{u,a},b});
  }
  vec c(n),d(n);
  for(ll i=0;i<n;i++) cin >> c[i] >> d[i];
  priority_queue<ppll,vector<ppll>,greater<ppll>> q;
  q.push({{0,0},min(5000LL,s)});
  mat dp(n,vec(5001,-1));
  vec ans(n);
  while(q.size()) {
    auto p=q.top();
    q.pop();
    ll f=p.first.second;
    ll dis=p.first.first;
    ll re=p.second;
    if(dp[f][re]>-1) continue;
    dp[f][re]=dis;
    if(!ans[f]) ans[f]=dis;
    for(auto pp:G[f]) {
      ll co=pp.first.second;
      ll t=pp.first.first;
      ll nd=pp.second;
      if(re>=co&&dp[t][re-co]==-1) {
        q.push({{dis+nd,t},re-co});
      }
    }
    if(re+c[f]<=5000) q.push({{dis+d[f],f},re+c[f]});
  }
  for(ll i=1;i<n;i++) cout << ans[i] << endl;
}