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;

const int nm = 50, am = 50;
int n, m, s, r;
long long ans[nm], reach[nm][(nm - 1) * am + 1];
bool visited[nm];

struct Edge {
	int opp;
	int cost;
	int time;
	Edge(int o, int c, int t) :
			opp(o), cost(c), time(t) {
	}
};
vector<Edge> g[nm];

struct State {
	long long time;
	int num;
	int coin;
	State(long long t, int n, int c) :
			time(t), num(n), coin(c) {
	}
};
auto comp = [](State &a, State &b) {
	return a.time > b.time;
};

priority_queue<State, vector<State>, decltype(comp)> q(comp);

int main() {
	cin >> n >> m >> s;
	fill(reach[0], reach[n], LLONG_MAX);
	s = min(s, (nm - 1) * am);
	r = n - 1;
	for (int i = 0, u, v, a, b; i < m; i++) {
		cin >> u >> v >> a >> b;
		u--, v--;
		g[u].emplace_back(v, a, b);
		g[v].emplace_back(u, a, b);
	}
	for (int i = 0, c, d; i < n; i++) {
		cin >> c >> d;
		g[i].emplace_back(i, -c, d);
	}
	visited[0] = 1;
	for (int i = 0; i <= s; i++)
		reach[0][i] = 0;
	q.emplace(0, 0, s);
	while (1) {
		long long t = q.top().time;
		int v = q.top().num;
		int c = q.top().coin;
		q.pop();
		if (!visited[v]) {
			visited[v] = 1;
			ans[v] = t;
			if (!--r)
				break;
		}
		for (Edge &e : g[v]) {
			if (e.cost > c)
				continue;
			long long nt = t + e.time;
			int nv = e.opp;
			int nc = min(c - e.cost, (nm - 1) * am);
			if (nt < reach[nv][nc]) {
				for (int i = 0; i <= nc; i++)
					reach[nv][i] = min(reach[nv][i], nt);
				q.emplace(nt, nv, nc);
			}
		}

	}
	for (int i = 1; i < n; i++)
		cout << ans[i] << endl;

}
