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

struct edge { int dst, silver, time; };

int main() {
    int n, m, S; cin >> n >> m >> S;
    S = min(S, 2450);
    vector<edge> adj[n];
    for (int i = 0; i < m; i++) {
        int u, v, a, b; cin >> u >> v >> a >> b;
        u--; v--;
        adj[u].push_back({v, -a, b});
        adj[v].push_back({u, -a, b});
    }
    for (int i = 0; i < n; i++) {
        int c, d; cin >> c >> d;
        adj[i].push_back({i, c, d});
    }

    priority_queue<pair<long long, pair<int, int> > > pq;
    long long time[n][2451];
    for (int i = 0; i < n; i++) fill(time[i], time[i]+2451, 1LL<<60);
    pq.push({0, {0, S}});
    time[0][S] = 0;
    while (!pq.empty()) {
        long long t = -pq.top().first;
        int u = pq.top().second.first, s = pq.top().second.second;
        pq.pop();
        if (t > time[u][s]) continue;
        for (auto e : adj[u]) {
            long long t1 = t + e.time;
            int u1 = e.dst, s1 = min(s + e.silver, 2450);
            if (0 <= s1 && time[u1][s1] > t1) {
                time[u1][s1] = t1;
                pq.push({-t1, {u1, s1}});
            }
        }
    }
    for (int u = 1; u < n; u++)
        cout << *min_element(time[u], time[u] + 2451) << endl;
}
