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;

int main() {
    int N, M;
    ll S;
    cin >> N >> M >> S;
    ll C = 3000;
    vector<vector<ll> > dp(N, vector<ll>(C + 1, LLONG_MAX >> 4));

    vector<vector<tuple<int, int, ll>>> E(N);
    vector<pair<ll, ll>> B(N);

    for(int i = 0; i < M; i++) {
        int u, v, a;
        ll b;
        cin >> u >> v >> a >> b;
        E[u-1].push_back(make_tuple(v-1,a,b));
        E[v-1].push_back(make_tuple(u-1,a,b));
    }

    for(auto & [c, d]: B) cin >> c >> d;

    for(int i = 0; i <= min(C, S); i++) dp[0][i] = 0;
    for(int i = 0; i < 2000; i++) {
        for(int j = 0; j < N; j++) {
            for(ll s = 0; s <= C; s++) {
                for(auto [from, cost, t]: E[j]) {
                    if(s + cost > C) continue;
                    dp[j][s] = min(dp[j][s], dp[from][s + cost] + t);
                }
                ll b = dp[j][max(0LL, s - B[j].first)] + B[j].second;
                dp[j][s] = min(dp[j][s], b);
            }
        }
    }

    for(int i = 1; i < N; i++) cout << dp[i][0] << endl;
}