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;

struct Edge{
	int v, a, b;
};

const int MAX = 2500;
int n, m, s;
int c[51], d[51];
long long dist[51][2510];
vector<Edge> adj[51];
priority_queue<pair<long long, pair<int, int> > > pq;
bool vis[51][2510];

void Dijkstra(){
	for(int i = 1; i <= n; i++)
		for(int j = 0; j <= MAX; j++)
			dist[i][j] = 1e18;

	dist[1][min(MAX, s)] = 0;
	pq.push({0, {1, min(MAX, s)}});

	int cnt = 0;
	while(!pq.empty()){
		long long dis; int t, u;

		dis = -pq.top().first; tie(u, t) = pq.top().second; pq.pop();

		if(vis[u][t]) continue;

		vis[u][t] = true;

		for(auto &k : adj[u]){
			if(t - k.a >= 0 && dist[k.v][t - k.a] > dis + k.b){
				++cnt;
				dist[k.v][t - k.a] = dis + k.b;
				pq.push({-dist[k.v][t - k.a], {k.v, t - k.a}});
			}
		}

		long long tp = 0;
		while(true){
			t += c[u];
			tp += d[u];

			if(dist[u][min(MAX, t)] > dis + tp){
				dist[u][min(MAX, t)] = dis + tp;
				pq.push({-(dis + tp), {u, min(MAX, t)}});
			} else break;

			if(t > MAX) break;
		}
	}

	// cout << cnt << "\n";
}

int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

	cin >> n >> m >> s;

	while(m--){
		int u, v, a, b;

		cin >> u >> v >> a >> b;

		adj[u].emplace_back(Edge({v, a, b}));
		adj[v].emplace_back(Edge({u, a, b}));
	}

	for(int i = 1; i <= n; i++) cin >> c[i] >> d[i];

	Dijkstra();

	for(int i = 2; i <= n; i++){
		long long ans = 1e18;
		for(int sil = 0; sil <= MAX; sil++){
			ans = min(ans, dist[i][sil]);
		}

		cout << ans << "\n";
	}
	
	return 0;
}