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 <iostream>
#include <algorithm>
#include <functional>
#include <vector>
#include <queue>
#include <tuple>
#define int long long
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

struct Edge {
	int to, mai, t;
	Edge() {}
	Edge(int to, int mai, int t) {
		this->to = to;
		this->mai = mai;
		this->t = t;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

int INF = 1e+15;
int n, m, s;
Graph g;
int c[50], d[50];
int dp[50][2501];

signed main() {
	int i, j;
	
	cin >> n >> m >> s;
	
	g.resize(n);
	rep(i, m) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		u--; v--;
		g[u].push_back(Edge(v, a, b));
		g[v].push_back(Edge(u, a, b));
	}
	rep(i, n) cin >> c[i] >> d[i];
	
	rep(i, n) rep(j, 2501) dp[i][j] = INF;
	
	typedef tuple<int, int, int> T;	//(time, pos, maisu)
	priority_queue<T, vector<T>, greater<T>> que;
	
	s = min(s, (int)2500);
	que.push(T(0, 0, s));
	
	while (!que.empty()) {
		T now = que.top();
		que.pop();
		
		int t = get<0>(now);
		int pos = get<1>(now);
		int mai = get<2>(now);
		if (dp[pos][mai] <= t) continue;
		dp[pos][mai] = t;
		
		//移動
		rep(i, g[pos].size()) {
			Edge e = g[pos][i];
			if (mai >= e.mai) {
				que.push(T(t + e.t, e.to, mai - e.mai));
			}
		}
		
		//換金
		que.push(T(t + d[pos], pos, min(mai + c[pos], (int)2500)));
	}
	
	for (i = 1; i < n; i++) {
		int ans = INF;
		rep(j, 2501) {
			ans = min(ans, dp[i][j]);
		}
		cout << ans << endl;
	}
	return 0;
}