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 int64_t ll;
using P = pair<ll, ll>;
#define MAX_V 122551
#define MAX_C 2450
#define INF 1e16
vector<vector<P>> Graph(MAX_V,vector<P>());
vector<ll> D(MAX_V,INF);

void dijkstra(int s) {
  priority_queue<P, vector<P>, greater<P>> que;
  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 (int i=0; i<(int)Graph[v].size(); ++i) {
      P e = Graph[v][i];
      if (D[e.first] > D[v] + e.second) {
        D[e.first] = D[v] + e.second;
        que.push(P(D[e.first], e.first));
      }
    }
  }
}

int main() {
  int n,m,s;
  cin >> n >> m >> s;
  if(s>MAX_C) s=MAX_C;
  int u,v,a,b;
  for(int i=0;i<m;i++){
    cin >> u >> v >> a >> b;
    for(int j=0;j<=MAX_C;j++){
      if (j-a >= 0){
        Graph[50*j+u].push_back(P(50*(j-a)+v,b));
        Graph[50*j+v].push_back(P(50*(j-a)+u,b));
      }
    }
  }
  int c,d,tmp;
  for(int i=1;i<=n;i++){
    cin >> c >> d;
    for(int j=0;j<=MAX_C;j++){
      tmp = j+c;
      if(tmp > MAX_C) tmp = MAX_C;
      Graph[50*j+i].push_back(P(50*tmp+i,d));
    }
  }
  dijkstra(50*s+1);
  ll mn;
  for(int i=2;i<=n;i++){
    mn = INF;
    for(int j=0;j<=MAX_C;j++){
      mn = min(mn,D[50*j+i]);
    }
    cout << mn << endl;
  }
}
