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 <cstdio>
#include <set>
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;

#define ll long long
const ll N = 1000000000000000;

int main(){
    int n, m, s;
    cin >> n >> m >> s;
    vector<ll> ans(n, N);

    vector<vector< vector<int> > > gr(n);
    for (int i = 0; i < m; ++i) {
        int u, v, a, b;
        cin >> u >> v >> a >> b; --u;--v;
        vector<int> state = {v, a, b};
        gr[u].emplace_back(state);
        state[0] = u;
        gr[v].emplace_back(state);
    }

    int CC = 2600;
    ll dp[n][CC];
    for(int i = 0; i < n; ++i) {
        fill(dp[i], dp[i] + CC, N);
        int a, b;
        cin >> a >> b;
        vector<int> state = {i, -a, b};
        gr[i].emplace_back(state);
    }

    s = min(2500, s);
    vector<ll> state = {0, s, 0};
    queue<vector<ll> > q;
    q.push(state);
    while(!q.empty()) {
        int old = q.size();
        while (old--) {
            state = q.front();
            q.pop();
            for (auto i : gr[state[0]]) {
                auto nst = state;
                nst[0] = i[0];
                nst[1] -= i[1];
                nst[2] += i[2];
                if (nst[1] < 0 || nst[1] > CC - 1 || dp[nst[0]][nst[1]] <= nst[2]) continue;
                dp[nst[0]][nst[1]] = nst[2];
                q.push(nst);
            }
        }
    }

    for (int i = 1; i < n; ++i) {
        ll mn = N;
        for (ll j : dp[i]) mn = min(j, mn);
        cout << mn << endl;
    }
    return 0;
}