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;

#define LL long long
#define ULL unsigned LL
#define LD long double
#define debug(x) cerr << #x << " " << x << endl;

const LL INF = 1e18 + 2137;
const int N = 51 * 5002 + 2137;
vector<pair<int, LL>> V[N];
LL odl[N];

void djikstra(int src) {
	odl[src] = 0;
	set<pair<LL, int>> S;
	S.insert(make_pair((LL) 0, src));
	while(!S.empty()) {
		pair<LL, int> tmp = *(S.begin());
		S.erase(S.begin());
		int v = tmp.second;
		for(int i=0; i<V[v].size(); i++) {
			int u = V[v][i].first;
			LL waga = V[v][i].second;
			if(odl[u] > odl[v] + waga) {
				if(odl[u] != INF)
					S.erase(S.find(make_pair(odl[u], u)));
				odl[u] = odl[v] + waga;
				S.insert(make_pair(odl[u], u));
			}
		}
	}
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int n, m, s;
	cin>>n>>m>>s;
	s = min(s, 5000);
	int u, v, a, b;
	for(int i=0; i<m; i++) {
		cin>>u>>v>>a>>b;
		for(int j=a; j<=5000; j++) {
			V[j*n+u].push_back({(j-a)*n+v, b});
			V[j*n+v].push_back({(j-a)*n+u, b});
		}
	}
	int c, d;
	for(int i=1; i<=n; i++) {
		cin>>c>>d;
		for(int j=0; (j+c) <= 5000; j++) {
			V[j*n+i].push_back({(j+c)*n+i, d});
		}
	}
	for(int i=1; i<=n; i++) {
		for(int j=0; j<=5000; j++)
			odl[j*n+i] = INF;
	}
	int nrstartowy = s*n + 1;
	djikstra(nrstartowy);
	for(int i=2; i<=n; i++) {
		LL odlx = INF;
		for(int j=0; j<=5000; j++) {
			odlx = min(odlx, odl[j*n+i]);
		}
		cout<<odlx<<"\n";
	}
}
