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 N = 300005;
const int S = 5000;
const long long inf = 1e18;

int num(int v, int s) {
	return v * (S + 1) + s;
}

int n, m, s, x, y, z, w;
long long d[N];
set<pair<long long, int> > se;
vector<pair<int, int> > v[N];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m >> s;
	s = min(s, S);
	for(int i = 1; i <= m; ++i) {
		cin >> x >> y >> z >> w;
		for(int j = z; j <= S; ++j) {
			v[num(x, j)].push_back({num(y, j - z), w});
			v[num(y, j)].push_back({num(x, j - z), w});
		}
	}
	for(int i = 1; i <= n; ++i) {
		cin >> x >> y;
		for(int j = 0; j <= S - x; ++j) {
			v[num(i, j)].push_back({num(i, j + x), y});
		}
		for(int j = 1; j <= S; ++j) {
			v[num(i, j)].push_back({num(i, j - 1), 0});
		}
	}
	for(int i = 0; i < N; ++i) {
		d[i] = inf;
	}
	se.insert({0, num(1, s)});
	d[num(1, s)] = 0;
	while(se.size()) {
		int dg = (*se.begin()).second;
		se.erase(se.begin());
		for(int i = 0; i < v[dg].size(); ++i) {
			int nv = v[dg][i].first, mz = v[dg][i].second;
			if(d[nv] > d[dg] + mz) {
				if(se.find({d[nv], nv}) != se.end()) {
					se.erase(se.find({d[nv], nv}));
				}
				d[nv] = d[dg] + mz;
				se.insert({d[nv], nv});
			}
		}
	}
	for(int i = 2; i <= n; ++i) {
		cout << d[num(i, 0)] << "\n";
	}
	cout << flush;
	return 0;
}
