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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
struct edge{ll to; ll cost;};
typedef pair<ll,ll> P;
const ll INF = 1e17;
ll MAX = 3000;

void dijkstra(int n, int s, vector<vector<edge>>& g, vector<ll>& d){
  priority_queue<P,vector<P>,greater<P>> que;
  rep(i,n) d[i] = INF;
  d[s] = 0;
  que.push(P(0,s));
  while(!que.empty()){
    P p = que.top();
    que.pop();
    int v = p.second;
    if(d[v] < p.first) continue;
    for(auto e : g[v]){
      if(d[e.to] > d[v] + e.cost){
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main(){
  ll n, m, s;
  cin >> n >> m >> s;
  ll u[m], v[m], a[m], b[m];
  rep(i,m) {
    cin >> u[i] >> v[i] >> a[i] >> b[i];
    u[i]--; v[i]--;
  }
  ll c[n], d[n];
  rep(i,n) cin >> c[i] >> d[i];
  vector<vector<edge>> g(n*MAX);
  vector<ll> dist(n*MAX);
  for(int i = 0; i < n; i++) {
    for(int j = 0; j < MAX; j++) {
      g[i*MAX+j].push_back(edge{i*MAX+min(MAX-1,c[i]+j),d[i]});
    }
  }
  rep(i,m) {
    rep(j,MAX) {
      if(j-a[i] >= 0) {
        g[u[i]*MAX+j].push_back(edge{v[i]*MAX+j-a[i],b[i]});
        g[v[i]*MAX+j].push_back(edge{u[i]*MAX+j-a[i],b[i]});
      }
    }
  }
  s = min(MAX-1,s);
  dijkstra(n*MAX,s,g,dist);
  for(int i = 1; i < n; i++) {
    ll ans = INF;
    rep(j,MAX) ans = min(ans,dist[i*MAX+j]);
    cout << ans << endl;
  }
  return 0;
}
