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;

using ll = long long;
const int MAX = 50;
const int LAYERS = 2500;

int main() {
    int N, M, S;
    cin >> N >> M >> S;
    vector<vector<tuple<int, int, int>>> adj(N);
    for (int i = 0; i < M; i++) {
        int u, v, a, b;
        cin >> u >> v >> a >> b, u--, v--;
        adj[u].emplace_back(v, a, b);
        adj[v].emplace_back(u, a, b);
    }
    vector<int> Cs(N), Ds(N);
    for (int i = 0; i < N; i++) {
        cin >> Cs[i] >> Ds[i];
    }
    vector<ll> ans(N);
    set<int> has, vis;
    priority_queue<pair<ll, int>> pq;
    pq.emplace(0, min(LAYERS, S) * MAX);
    while (has.size() < N) {
        ll w;
        int ind, n, s;
        tie(w, ind) = pq.top();
        pq.pop();
        if (vis.count(ind))
            continue;
        vis.emplace(ind);
        s = ind / MAX;
        n = ind % MAX;
        if (!has.count(n)) {
            has.emplace(n);
            ans[n] = w;
        }
        pq.emplace(w - Ds[n], n + min(LAYERS, s + Cs[n]) * MAX);
        for (auto edge : adj[n]) {
            int v, a, b;
            tie(v, a, b) = edge;
            if (a > s)
                continue;
            int nind = v + min(LAYERS, s - a) * MAX;
            if (vis.count(nind))
                continue;
            pq.emplace(w - b, nind);
        }
    }
    for (int i = 1; i < N; i++) {
        cout << -ans[i] << endl;
    }
}
