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 int MAX_V = 50;
const int MAX_S = MAX_V * 50 + 5;
const long long INF = 1e18;

struct Edge {
	int to, a, b;
	Edge(int to, int a, int b) :to(to), a(a), b(b) {}
};

struct Data {
	int v, s;//頂点、所持金
	long long x;//時間
	Data(int v, int s, long long x) :v(v), s(s), x(x) {}
	bool operator<(const Data& a) const {//小さい順
		return x > a.x;
	}
};

vector<Edge>g[MAX_V];
long long dp[MAX_V][MAX_S + 5];
int main() {
	//dp[頂点][金]=時間
	int n, m, s;
	cin >>n>>m>>s;
	for (int i = 0;i<m;++i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		--u;
		--v;
		g[u].emplace_back(v, a, b);
		g[v].emplace_back(u, a, b);
	}

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

	for (int i = 0;i < MAX_V;++i) {
		for (int j = 0;j<MAX_S+5;++j) {
			dp[i][j]= INF;
		}
	}
	s = min(s, MAX_S);

	priority_queue<Data>q;
	auto push = [&](int v, int s,long long x) {
		if (s < 0) {
			return;
		}
		if (dp[v][s] <= x) {
			return;
		}
		dp[v][s] = x;
		q.emplace(v, s, x);
	};
	push(0,s,0);//初期条件

	while (!q.empty()) {
		Data hoge = q.top();
		q.pop();
		int v = hoge.v;
		int s = hoge.s;
		long long x = hoge.x;
		if (dp[v][s] != x) {
			continue;
		}
		//お金増やす
		int ms = min(s+c[v],MAX_S);
		push(v, ms, x + d[v]);
		//移動
		for (Edge e:g[v]) {
			push(e.to, s - e.a, x + e.b);
		}
	}
	for (int i = 1;i<n;++i) {
		long long  ans = INF;
		for (int j = 0;j < MAX_S + 5; ++j) {
			ans = min(ans, dp[i][j]);
		}
		cout << ans << endl;
	}

	return 0;
}