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 int ll;
typedef unsigned long long int ull;
typedef pair<ll, ll> PLL;
#define _overload3(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) \
  for (int i = static_cast<int>(a); i < static_cast<int>(b); ++i)
#define rep(...) _overload3(__VA_ARGS__, repi, _rep, ) (__VA_ARGS__)  // NOLINT

ll N, M, S;
struct Edge {
	ll u, a, b;
};
vector<Edge> edges[60];
const ll INF = 1e18;
ll C[60], D[60];
ll dp[60][20000];
typedef pair<ll, pair<ll, ll>> P;

signed main() {
	cin >> N >> M >> S;
	S = min(S, 19000ll);
	rep(i, M) {
		ll u, v, a, b;
		cin >> u >> v >> a >> b;
		edges[u].push_back({v, a, b});
		edges[v].push_back({u, a, b});
	}
	for (int i=1; i<=N; i++) {
		cin >> C[i] >> D[i];
	}
	rep(i, 60) rep(j, 20000) dp[i][j] = INF;
	priority_queue<P, vector<P>, std::greater<P>> q; // {time, {vertex, money}}
	q.push({0ll, {1ll, S}});
	while (!q.empty()) {
		auto g = q.top();
		q.pop();
		ll t = g.first, i = g.second.first, j = g.second.second;
		if (dp[i][j] <= t)
			continue;
		dp[i][j] = t;
		// printf("i=%lld j=%lld t=%lld dp[i][j]=%lld\n", i, j, t, dp[i][j]);
		// exchange
		if (j + C[i] < 20000)
			q.push({t + D[i], {i, j + C[i]}});
		// move
		for (auto e: edges[i]) {
			ll u = e.u, a = e.a, b = e.b;
			if (j-a >= 0)
				q.push({t+b, {u, j-a}});
		}
	}
	for (int i=2; i<=N; i++) {
		ll ans = INF;
		rep(j, 20000)
			ans = min(ans, dp[i][j]);
		cout << ans << endl;
	}
	return 0;
}
