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 ll;
int main() {
	int N, M, S;
	cin >> N >> M >> S;
	vector<vector<int>> G(N);
	map<pair<int,int>,int> A;
	map<pair<int,int>,int> B;
	for (int i = 0; i < M; ++ i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		-- u; -- v;
		G[u].push_back(v);
		G[v].push_back(u);
		A[{u,v}] = a;
		A[{v,u}] = a;
		B[{u,v}] = b;
		B[{v,u}] = b;
	}
	vector<int> C(N), D(N);
	for (int i = 0; i < N; ++ i) cin >> C[i] >> D[i];

	vector<vector<ll>> dist(50, vector<ll>(4001, 1LL<<62));
	priority_queue<pair<ll,pair<int,int>>> Q;

	int ss = min(S, 4000);
	dist[0][ss] = 0;
	Q.push({0, {0,ss}});
	vector<ll> res(N, 1LL<<62);
	while (!Q.empty()) {
		ll d = -Q.top().first;
		int u = Q.top().second.first;
		int s = Q.top().second.second;
		Q.pop();
		res[u] = min(res[u], d);
		if (d != dist[u][s]) continue;
		if (s < 4000) {
			int ss = min(s + C[u], 4000);
			ll dd = d + D[u];
			if (dd < dist[u][ss]) {
				dist[u][ss] = dd;
				Q.push({-dd,{u,ss}});
			}
		}
		for (auto x : G[u]) {
			int ss = s - A[{u,x}];
			if (ss >= 0) {
				ll dd = d + B[{u,x}];
				if (dd < dist[x][ss]) {
					dist[x][ss] = dd;
					Q.push({-dd,{x,ss}});
				}
			}
		}
	}
	for (int i = 1; i < N; ++ i) cout << res[i] << endl;
}
