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;

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n, m, s; cin >> n >> m >> s;
    using edge = array<int, 3>;
    vector<vector<edge>> e(n);
    vector<int> c(n), d(n);
    for (int i = 0; i < m; i++) {
        int u, v, a, b; cin >> u >> v >> a >> b;
        u--, v--;
        e[u].push_back({v, a, b});
        e[v].push_back({u, a, b});
    }
    for (int i = 0; i < n; i++) cin >> c[i] >> d[i];
    const int A = n * 50;
    const long long inf = 1e18;
    long long cost[n][A+1];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= A; j++) {
            cost[i][j] = inf;
        }
    }
    s = min(A, s);
    cost[0][s] = 0;
    set<tuple<long long, int, int>> st;
    st.emplace(cost[0][s], 0, s);
    while (!st.empty()) {
        auto [x, i, ai] = *st.begin();
        st.erase(st.begin());
        if (cost[i][ai] != x) continue;
        if (cost[i][min(A, ai+c[i])] > x + d[i]) {
            cost[i][min(A, ai+c[i])] = x + d[i];
            st.emplace(x+d[i], i, min(A, ai+c[i]));
        }
        for (auto &[j, a, b]: e[i]) {
            if (ai < a) continue;
            if (cost[j][ai-a] > x + b) {
                cost[j][ai-a] = x + b;
                st.emplace(x+b, j, ai-a);
            }
        }
    }
    for (int i = 1; i < n; i++) {
        long long ans = inf;
        for (int a = 0; a <= A; a++) {
            ans = min(ans, cost[i][a]);
        }
        cout << ans << "\n";
    }
    
    return 0;
}