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>
#define Int int64_t

using namespace std;

void chmin(Int& a, Int b) { if (a > b) { a = b; } };

int main() {
	constexpr Int INF = 1e18;
	constexpr Int maxS = 2500;
	Int N, M, S;
	cin >> N >> M >> S;
	vector<vector<tuple<int, int, Int>>> g(N);
	for (int i = 0; i < M; ++i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		--u; --v;
		g[u].emplace_back(v, a, b);
		g[v].emplace_back(u, a, b);
	}
	vector<Int> C(N), D(N);
	for (int i = 0; i < N; ++i) {
		cin >> C[i] >> D[i];
	}

	vector<vector<Int>> d(N, vector<Int>(maxS + 1, INF));
	d[0][min(S, maxS)] = 0;
	priority_queue<pair<Int, int>, vector<pair<Int, int>>, greater<pair<Int, int>>> que;
	que.emplace(0, min(S, maxS));
	while (!que.empty()) {
		auto p = que.top(); que.pop();
		Int cost = p.first;
		int u = p.second / 10000;
		int s = p.second % 10000;

		if (s < maxS) {
			int nxs = min(maxS, s + C[u]);
			if (d[u][nxs] > d[u][s] + D[u]) {
				d[u][nxs] = d[u][s] + D[u];
				que.emplace(d[u][nxs], u*10000 + nxs);
			}
		}
		for (auto t : g[u]) {
			int v = get<0>(t);
			int a = get<1>(t);
			Int b = get<2>(t);

			if (s - a < 0 || d[v][s - a] <= d[u][s] + b) {
				continue;
			}
			d[v][s - a] = d[u][s] + b;
			que.emplace(d[v][s - a], v*10000 + (s - a));
		}
	}

	for (int i = 1; i < N; ++i) {
		Int ans = INF;
		for (auto val : d[i]) { ans = min(ans, val); }
		cout << ans << endl;
	}

	return 0;
}
