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 <utility>
#include <queue>
#include <functional>
#include <algorithm>

using namespace std;

typedef pair<int, int> pii;
typedef vector<pii> vp;
typedef long long ll;
typedef pair<ll, int> pli;
typedef vector<pli> vpli;
typedef priority_queue<pli, vpli, greater<pli> > pqp;

const int A = 50;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m, s;
    cin >> n >> m >> s;
    int p = 2 * n * A + 1;
    vp e[n*p];
    for (int i = 0; i < m; i++) {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        u--, v--;
        for (int j = 0; j < p - a; j++) {
            e[u*p+j+a].push_back({b, v*p+j});
            e[v*p+j+a].push_back({b, u*p+j});
        }
    }
    for (int i = 0; i < n; i++) {
        int c, d;
        cin >> c >> d;
        for (int j = 0; j < p; j++) {
            e[i*p+j].push_back({d, i*p+min(p-1,j+c)});
        }
    }

    s = min(s, p - 1);

    ll dis[n*p];
    fill(dis, dis+n*p, 3e12);

    pqp q;
    q.push({0, s});
    dis[s] = 0;

    while (!q.empty()) {
        ll d = q.top().first;
        int x = q.top().second;
        q.pop();

        if (d != dis[x]) {
            continue;
        }

        for (pii y : e[x]) {
            if (d + y.first < dis[y.second]) {
                dis[y.second] = d + y.first;
                q.push({dis[y.second], y.second});
            }
        }
    }

    for (int i = 1; i < n; i++) {
        cout << *min_element(dis+i*p, dis+(i+1)*p) << endl;
    }

    return 0;
}
