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;
using lint = long long;
const lint INF = 1e18, MAX_COST = 2500;
typedef pair<lint, lint> P;
using edge = pair<lint, P>;

signed main(){
	lint N, M, S; cin >> N >> M >> S;
	S = min(S, MAX_COST);
	vector<vector<edge> > G(N);
	for(lint i = 0; i < M; i++){
		lint u, v, a, b; cin >> u >> v >> a >> b; --u; --v;
		G[u].push_back(edge(v, P(a, b))); G[v].push_back(edge(u, P(a, b)));
	}
	vector<lint> c(N), d(N);
	for(lint i = 0; i < N; i++) cin >> c[i] >> d[i];
	vector<vector<lint> > dist(N, vector<lint> (MAX_COST + 1, INF));
	priority_queue<edge, vector<edge>, greater<edge> > que;
	dist[0][S] = 0;
	que.push(edge(0, P(0, S)));

	while(que.size()){
		edge e = que.top(); que.pop();
		lint time = e.first, now = e.second.first, money = e.second.second;
		if(dist[now][money] < time) continue;

		if(money + c[now] <= MAX_COST){
			lint ntime = time + d[now], nmoney = money + c[now];
			if(dist[now][nmoney] > ntime){
				dist[now][nmoney] = ntime;
				que.push(edge(ntime, P(now, nmoney))); 
			}
		}

		for(auto e : G[now]){
			if(money < e.second.first) continue;
			lint next = e.first, ntime = time + e.second.second, nmoney = money - e.second.first;
			if(dist[next][nmoney] > ntime){
				dist[next][nmoney] = ntime;
				que.push(edge(ntime, P(next, nmoney)));
			}
		}
	}

	for(lint i = 1; i < N; i++){
		lint ans = INF;
		for(lint j = 0; j <= MAX_COST; j++){
			ans = min(ans, dist[i][j]);
		}
		cout << ans << endl;
	}
}