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<(n); ++i)
typedef vector<int> vi;
typedef long long int ll;
#define LL_MAX LLONG_MAX


//DIJKSTRA
struct edge {
    int to;
    int cost;
};

typedef pair<ll, int> P;  // first: 最短距離, second: 頂点番号

ll V, d[50*5000];
vector<edge> G[50*5000]; // 隣接リスト表現

void dijkstra(int s) {
    fill(d, d + V, LL_MAX);
    d[s] = 0;  // 始点sへの最短距離は0

    priority_queue<P, vector<P>, greater<P>> que;  // 距離が小さい順に取り出せるようgreater<P>を指定
    que.push(P(0, s));
    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second;  // 更新した頂点の中で距離が最小の頂点v
        if (d[v] < p.first) {
            continue;
        }
        for (auto e : G[v]) {  // 頂点vから出る辺eを走査
            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() {
  int n, m, s;
  cin >> n >> m >> s;
  s = min(s, 4999);
  V = n * 5000;
  int u, v, a, b;
  rep(i,m) {
    cin >> u >> v >> a >> b;
    --u; --v;
    for (int i=a; i<5000; ++i) {
      G[u*5000+i].push_back({v*5000+i-a, b});
      G[v*5000+i].push_back({u*5000+i-a, b});
    }
  }
  int c, dd;
  rep(i,n){
    cin >> c >> dd;
    for(int j=0; j<5000-c; ++j) {
      G[i*5000+j].push_back({i*5000+j+c, dd});
    }
  }
  dijkstra(s);
  for(int i=1;i<n;++i){
    ll ans = LL_MAX;
    rep(j,5000) ans = min(ans, d[i*5000+j]);
    cout << ans << "\n";
  }
  return 0;
}
