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;

typedef long long ll;
typedef tuple<ll, ll, ll> tlll;
typedef pair<ll, ll> Pll;

ll N, M, S;
vector<vector<tlll>> graph;
vector<Pll> excha;

constexpr ll INF = 1e14;
constexpr ll MAXCOST = 2600;

int main() {
    cin >> N >> M >> S;
    graph.resize(N);
    for (int i = 0; i < M; i++) {
        ll U, V, A, B; cin >> U >> V >> A >> B;
        U--; V--;
        graph[U].push_back(tlll(V, A, B));
        graph[V].push_back(tlll(U, A, B));
    }
    excha.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> excha[i].first >> excha[i].second;
    }

    vector<vector<ll>> ans(N, vector<ll>(MAXCOST+1, INF));
    // time, node, money;
    priority_queue<tlll, vector<tlll>, greater<tlll>> pq;
    pq.push(tlll(0, 0, min(S, MAXCOST)));

    while (!pq.empty()) {
        ll t, n, m; tie(t, n, m) = pq.top(); pq.pop();
        if (ans[n][m] != INF) continue;

        ans[n][m] = t;
        ll nxt = m + excha[n].first;
        ll tt = t + excha[n].second;
        if (nxt <= MAXCOST) {
            if (ans[n][nxt] > tt) pq.push(tlll(tt, n, nxt));
        }
        for (auto x : graph[n]) {
            ll u, a, b; tie(u, a, b) = x;
            if (m >= a) {
                if (ans[u][m-a] > t+b) pq.push(tlll(t+b, u, m-a));
            }
        }
    }
    for (int i = 1; i < N; i++) {
        ll a = INF;
        for (int j = 0; j <= MAXCOST; j++) {
            a = min(a, ans[i][j]);
        }
        cout << a << endl;
    }
}
