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

using edge = std::pair<int, int>;
std::vector<edge> edges[2501 * 50];
long long dist[2501 * 50];

int main()
{
	int n, m, s;
	scanf("%d%d%d", &n, &m, &s);
	for (int i = 0; i < m; ++i)
	{
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		u = (u - 1) * 2501;
		v = (v - 1) * 2501;
		for (int j = 0; a + j <= 2500; ++j)
		{
			edges[v + a + j].emplace_back(u + j, b);
			edges[u + a + j].emplace_back(v + j, b);
		}
	}
	for (int i = 0; i < n; ++i)
	{
		int c, d;
		scanf("%d%d", &c, &d);
		if (c > 2500)
		{
			c = 2500;
		}
		int v = i * 2501;
		for (int j = 0; j + c <= 2500; ++j)
		{
			edges[v + j].emplace_back(v + j + c, d);
		}
	}

	std::fill(dist, dist + 2501 * 50, 1e18);
	dist[std::min(2500, s)] = 0;
	using P = std::pair<long long, int>;
	std::priority_queue<P, std::vector<P>, std::greater<P>> q;
	q.emplace(0, std::min(2500, s));
	while (q.size() > 0)
	{
		auto [cost, v] = q.top(); q.pop();
		if (cost > dist[v])
		{
			continue;
		}
		for (auto [to, c] : edges[v])
		{
			long long next = cost + c;
			if (dist[to] > next)
			{
				dist[to] = next;
				q.emplace(next, to);
			}
		}
	}
	for (int i = 1; i < n; ++i)
	{
		long long ans = 1e18;
		for (int j = 0; j <= 2500; ++j)
		{
			if (ans > dist[i * 2501 + j])
			{
				ans = dist[i * 2501 + j];
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}
