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 <iostream>
#include <vector>
#include <queue>
using namespace std;

typedef long long ll;

typedef pair<int,int> P;
typedef pair<ll,P> PP;

struct edge{
  int to;
  int a;
  ll b;
};

int n,m;
ll s;
vector<edge> g[55];
ll c[55];
ll d[55];
ll di[55][2505];
bool use[55][2505];
const ll INF = 1e18;

int main() {
cin >> n >> m >> s;
s = min(2500ll,s);
for(int i = 0; i < m; i++) {
  int u,v,a,b;
  cin >> u >> v >> a >> b;
  u--,v--;
  g[u].push_back(edge{v,a,b});
  g[v].push_back(edge{u,a,b});
}

for(int i = 0; i < n; i++) {
  cin >> c[i] >> d[i];
}

for(int i = 0; i < n; i++) {
  for(int j = 0; j <= 2500; j++) {
    di[i][j] = INF;
  }
}

di[0][s] = 0;
priority_queue<PP, vector<PP>,greater<PP> > q;

q.push(PP(0,P(s,0)));

while(!q.empty()) {
  PP p = q.top();
  q.pop();
ll time = p.first;
int ginka = p.second.first;
int u = p.second.second;

if(di[u][ginka] < time) continue;



  if(di[u][min(2500ll,c[u] + ginka)] > di[u][ginka] + d[u]) {
    di[u][min(2500ll,c[u]+ginka)] = di[u][ginka] + d[u];
    q.push(PP(di[u][min(2500ll,c[u]+ginka)],P(min(2500ll,c[u]+ginka),u)));
  }

for(int i = 0; i < g[u].size(); i++) {
  edge e = g[u][i];

  if(ginka - e.a >= 0 && di[e.to][ginka-e.a] > di[u][ginka] + e.b) {
    di[e.to][ginka-e.a] = di[u][ginka] + e.b;
    q.push(PP(di[e.to][ginka-e.a],P(ginka-e.a,e.to)));
  }
}
}



for(int i = 1; i < n; i++) {
  ll ans = INF;
  for(int j = 0; j <= 2500; j++) {
    ans = min(ans,di[i][j]);
  
  }
  cout << ans << endl;
}




}