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;

long long n, m, s, u, v, z;
long long d[51][2500];
const long long INF = 1e18;

struct edge {
	long long to, c, tm;
};

vector<edge> G[51];
typedef tuple<long long, long long, long long> P;
priority_queue<P, vector<P>, greater<P>> Q;

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

	fill((long long*)d, (long long*)(d + n + 1), INF);

	for (int i = 0; i < m; i++) {
		edge e;
		cin >> u >> v >> e.c >> e.tm;

		e.to = v;
		G[u].push_back(e);
		e.to = u;
		G[v].push_back(e);

		z = max(z, e.c);
	}

	z *= n - 1;
	s = min(s, z);

	for (int i = 1; i <= n; i++) {
		long long c, d;
		cin >> c >> d;

		edge e;
		e.to = i;
		e.c = -c;
		e.tm = d;
		G[i].push_back(e);
	}

	d[1][s] = 0;
	Q.push(P(0, s, 1));

	while (!Q.empty()) {
		P p = Q.top();
		Q.pop();

		int ut = get<2>(p), uc = get<1>(p);

		if (d[ut][uc] < get<0>(p))
			continue;

		for (edge e : G[ut]) {
			long long uw = uc - e.c;

			uw = min(uw, z);

			if (uw < 0)
				continue;

			if (d[e.to][uw] > d[ut][uc] + e.tm) {
				d[e.to][uw] = d[ut][uc] + e.tm;
				Q.push(P(d[e.to][uw], uw, e.to));
			}
		}
	}

	for (int i = 2; i <= n; i++) {
		long long mn = INF;
		for (int j = 0; j <= z; j++)
			mn = min(mn, d[i][j]);

		cout << mn << endl;
	}
}
