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;

const long long inf = 1000000000000000000;

int main() {
	int n, m, s;
	cin >> n >> m >> s;

	auto id = [&] (int i, int j) {
		return i * 2501 + j;
	};

	vector<vector<pair<int, int> > > G;
	G.resize(id(n, 0));

	for(int i = 0; i < m; ++i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		
		--u;
		--v;

		for(int j = 2500; j >= a; --j) {
			G[id(u, j)].emplace_back(id(v, j - a), b);
			G[id(v, j)].emplace_back(id(u, j - a), b);
		}
	}

	for(int i = 0; i < n; ++i) {
		int c, d;
		cin >> c >> d;

		c = min(c, 2500);

		for(int j = 0; j < 2500; ++j) {
			G[id(i, j)].emplace_back(id(i, min(j + c, 2500)), d);
		}
	}

	priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > q;
	
	q.emplace(0, id(0, min(2500, s)));

	vector<long long> d;
	d.assign(id(n, 0), inf);

	d[id(0, min(2500, s))] = 0;

	while(!q.empty()) {
		auto o = q.top();
		q.pop();

		int u = o.second;

		if(d[u] < o.first) {
			continue;
		}

		for(auto t : G[u]) {
			int v = t.first;
			int w = t.second;

			if(d[u] + w < d[v]) {
				d[v] = d[u] + w;
				q.emplace(d[v], v);
			}
		}
	}

	for(int i = 1; i < n; ++i) {
		long long ans = inf;

		for(int j = 0; j <= 2500; ++j) {
			ans = min(ans, d[id(i, j)]);
		}

		cout << ans << '\n';
	}

	return 0;
}
