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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>

using namespace std;

template <typename T>
vector<T> dijkstra(int s, vector<vector<pair<int, T> > >& G) {
	const T INF = numeric_limits<T>::max();
	const int N = G.size();
	vector<T> cost(N, INF);
	priority_queue<pair<T, int>, vector<pair<T, int> >, greater<pair<T, int> > > pq;
	cost[s] = 0;
	pq.push(make_pair(0, s));
	while (!pq.empty()) {
		pair<T, int> p = pq.top(); pq.pop();
		int now = p.second;
		if (cost[now] < p.first) continue;
		for (auto next : G[now]) {
			int nv = next.first, nw = next.second;
			if (cost[nv] > cost[now] + nw) {
				cost[nv] = cost[now] + nw;
				pq.push(make_pair(cost[nv], nv));
			}
		}
	}
	return cost;
}

int main() {
	int n, m, s; cin >> n >> m >> s;
	vector<vector<pair<int, long long> > > G(3000 * n);
	for (int i = 0; i < m; ++i) {
		int u, v, a, b; cin >> u >> v >> a >> b;
		--u, --v;
		for (int j = a; j < 3000; ++j) {
			G[3000 * u + j].emplace_back(3000 * v + j - a, b);
			G[3000 * v + j].emplace_back(3000 * u + j - a, b);
		}
	}
	for (int i = 0; i < n; ++i) {
		int c, d; cin >> c >> d;
		for (int j = 0; j < 3000 - c; ++j) {
			G[3000 * i + j].emplace_back(3000 * i + j + c, d);
		}
	}
	auto ans = dijkstra(min(s, 2999), G);
	for (int i = 1; i < n; ++i) {
		long long res = 1e18;
		for (int j = 0; j < 3000; ++j) res = min(res, ans[3000 * i + j]);
		cout << res << endl;
	}

	return 0;
}