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 <vector>
#include <queue>
#include <tuple>
#include <algorithm>
using namespace std;

long long N, M, S;
long long X[1 << 18], Y[1 << 18], A[1 << 18], B[1 << 18];
long long C[1 << 18], D[1 << 18];
vector<pair<int, int>> G[109];

priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<tuple<long long, int, int>>> Q;
long long dp[59][2509];

int main() {
	cin >> N >> M >> S;
	for (int i = 1; i <= M; i++) cin >> X[i] >> Y[i] >> A[i] >> B[i];
	for (int i = 1; i <= N; i++) cin >> C[i] >> D[i];
	for (int i = 1; i <= M; i++) {
		G[X[i]].push_back(make_pair(Y[i], i));
		G[Y[i]].push_back(make_pair(X[i], i));
	}

	for (int i = 1; i <= N; i++) {
		for (int j = 0; j <= 2500; j++) dp[i][j] = (1LL << 60);
	}
	dp[1][min(2500LL, S)] = 0;
	Q.push(make_tuple(0, 1, min(2500LL, S)));

	while (!Q.empty()) {
		int a1 = get<1>(Q.top());
		int a2 = get<2>(Q.top()); Q.pop();
		if (dp[a1][min(2500LL, a2 + C[a1])] > dp[a1][a2] + D[a1]) {
			dp[a1][min(2500LL, a2 + C[a1])] = dp[a1][a2] + D[a1];
			Q.push(make_tuple(dp[a1][min(2500LL, a2 + C[a1])], a1, min(2500LL, a2 + C[a1])));
		}
		for (pair<int, int> i : G[a1]) {
			int to = i.first, idx = i.second;
			if (a2 - A[idx] >= 0 && dp[to][a2 - A[idx]] > dp[a1][a2] + B[idx]) {
				dp[to][a2 - A[idx]] = dp[a1][a2] + B[idx];
				Q.push(make_tuple(dp[to][a2 - A[idx]], to, a2 - A[idx]));
			}
		}
	}

	for (int i = 2; i <= N; i++) {
		long long ret = (1LL << 60);
		for (int j = 0; j <= 2500; j++) ret = min(ret, dp[i][j]);
		cout << ret << endl;
	}
	return 0;
}