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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;

int main() {
	int n, m, s;
	cin >> n >> m >> s;
	vector<vector<pair<int, int>>> cm(n * 2501);
	rep(i, m) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		u = (u - 1) * 2501; v = (v - 1) * 2501;
		rep(i, 2501) if (i >= a) {
			cm[u + i].emplace_back(v + i - a, b);
			cm[v + i].emplace_back(u + i - a, b);
		}
	}
	rep(i, n) {
		int c, d, vi = i * 2501;
		cin >> c >> d;
		rep(j, 2500) cm[vi + j].emplace_back(vi + min(j + c, 2500), d);
	}
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> q;
	vector<bool> visited(n * 2501);
	vector<ll> ans(n * 2501);
	q.emplace(0, min(s, 2500));
	while(!q.empty()) {
		auto p = q.top();
		q.pop();
		ll time = p.first;
		int dest = p.second;
		if (visited[dest]) continue;
		visited[dest] = true;
		ans[dest] = time;
		for(auto x: cm[dest]) {
			int nxt = x.first, len = x.second;
			q.emplace(time + len, nxt);
		}
	}
	for(int i = 1; i < n; i++) {
		ll res = LLONG_MAX;
		rep(j, 2501) {
			int vi = i * 2501 + j;
			if (visited[vi]) res = min(res, ans[vi]);
		}
		cout << res << endl;
	}
}
