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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

template<typename T> using pqueue_inv = std::priority_queue<T, std::vector<T>, std::greater<T> >;
int main() {
	int n = ri();
	int m = ri();
	int s = ri(); // initials
	struct Info {
		int to;
		int cost;
		int time;
	};
	std::vector<Info> hen[n];
	for (int i = 0; i < m; i++) {
		int a = ri() - 1;
		int b = ri() - 1;
		int c = ri();
		int d = ri();
		hen[a].push_back({b, c, d});
		hen[b].push_back({a, c, d});
	}
	int exscore[n], extime[n];
	for (int i = 0; i < n; i++) {
		exscore[i] = ri();
		extime[i] = ri();
	}
	
	// [n][2500]
	int64_t dist[n][2501];
	for (auto &i : dist) for (auto &j : i) j = 1000000000000000000;
	pqueue_inv<std::pair<int64_t, std::pair<int, int> > > que;
	que.push({dist[0][std::min(2500, s)] = 0, {0, std::min(2500, s)}});
	while (que.size()) {
		auto i = que.top();
		que.pop();
		int id = i.second.first;
		int left = i.second.second;
		if (dist[id][left] != i.first) continue;
		auto go = [&] (int new_id, int new_left, int cost) {
			if (dist[new_id][new_left] > i.first + cost)
				que.push({dist[new_id][new_left] = i.first + cost, {new_id, new_left}});
		};
		// top up
		go(id, std::min(2500, left + exscore[id]), extime[id]);
		for (auto j : hen[id]) {
			if (j.cost > left) continue;
			go(j.to, left - j.cost, j.time);
		}
	}
	for (int i = 1; i < n; i++) {
		int64_t res = 1000000000000000000;
		for (int j = 0; j <= 2500; j++) res = std::min(res, dist[i][j]);
		printf("%" PRId64 "\n", res);
	}
	
	return 0;
}
