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>
typedef long long ll;
using namespace std;
const ll MAX = 2500LL;
typedef pair <ll, ll> P;

int main() {
    cin.sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    ll n, m, s;
    cin >> n >> m >> s;
    s = min(s, MAX - 1);

    vector<vector<P>> edge(n * MAX);
    for (int i = 0; i < m; i++) {
        ll u, v, a, b;
        cin >> u >> v >> a >> b;
        u--; v--;
        for (int j = a; j < MAX; j++) {
            edge[u * MAX + j].push_back({v * MAX + j - a, b});
            edge[v * MAX + j].push_back({u * MAX + j - a, b});
        }
    }

    for (int i = 0; i < n; i++) {
        ll c, d;
        cin >> c >> d;
        for (int j = 0; j < MAX; j++) {
            edge[i * MAX + j].push_back({i * MAX + min(j + c, MAX - 1), d});
        }
    }

    vector <ll> ans(n * MAX, 1LL << 60);
    priority_queue <P, vector<P>, greater<P>> q;
    ans[s] = 0;
    q.push({0, s});

    while (!q.empty()) {
        P p = q.top();
        q.pop();
        ll time = p.first;
        int cur = p.second;
        if (ans[cur] < time) continue;
        for (P nxt : edge[cur]) {
            int next_node = nxt.first;
            ll total_time = time + nxt.second;
            if (ans[next_node] > total_time) {
                ans[next_node] = total_time;
                q.push({total_time, next_node});
            }
        }
    }

    for (int i = 1; i < n; i++) {
        ll mn = 1LL << 60;
        for (int j = 0; j < MAX; j++) mn = min(mn, ans[i * MAX + j]);
        cout << mn << '\n';
    }
    return 0;
}
