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>
#define long long long int
using namespace std;

// @author: pashka

const int MAX = 2501;

struct edge {
    int dest;
    long time;
};

int main() {
    ios::sync_with_stdio(false);

    int n, m, s;
    cin >> n >> m >> s;

    s = min(s, MAX - 1);
    int nn = n * MAX;
    vector<vector<edge>> g(nn);

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

    set<pair<int, int>> q;
    vector<long> d(nn, LLONG_MAX);
    s = s * n;
    d[s] = 0;
    q.insert({d[s], s});
    while (!q.empty()) {
        auto p = *q.begin();
        q.erase(p);
        int x = p.second;
        for (auto e : g[x]) {
            int y = e.dest;
            long dd = d[x] + e.time;
            if (dd < d[y]) {
                q.erase({d[y], y});
                d[y] = dd;
                q.insert({d[y], y});
            }
        }
    }
    for (int i = 1; i < n; i++) {
        long res = LLONG_MAX;
        for (int j = 0; j < MAX; j++) {
            res = min(res, d[j * n + i]);
        }
        cout << res << " ";
    }

    return 0;
}