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>
#define sz(c)       int(c.size())
#define rep(i,a,b)  for (int i=a; i<(b); ++i)
#define per(i,a,b)  for (int i=(b)-1; i>=(a); --i)
#define fore(c,...) for (auto __VA_ARGS__:(c))
using namespace std;
typedef long long ll;



int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	cout<<fixed<<setprecision(10);

	int N,M,S;
	cin>>N>>M>>S;
	vector<vector<array<int,3>>> adj(N);
	rep(i,0,M) {
		int u,v,a,b;
		cin>>u>>v>>a>>b;
		u--,v--;
		adj[u].push_back({v,a,b});
		adj[v].push_back({u,a,b});
	}

	vector<int> C(N),D(N);
	rep(i,0,N) cin>>C[i]>>D[i];

	const int maxs=2500;
	const ll inf=(ll)1e18;
	vector<array<ll,maxs+1>> dist(N);
	rep(i,0,N) fill(dist[i].begin(),dist[i].end(),inf);
	S=min(S,maxs);
	dist[0][S]=0;
	using state=tuple<ll,int,int>;
	priority_queue<state,vector<state>,greater<state>> q;
	q.emplace(0,0,S);
	vector<ll> res(N,inf);
	while (!q.empty()) {
		ll d;
		int v,s;
		tie(d,v,s)=q.top();
		q.pop();
		if (d!=dist[v][s]) continue;
		res[v]=min(res[v],d);

		// buy silver coins
		for (ll i=1; s+i*C[v]<=maxs; ++i) {
			int ns=(int)(s+i*C[v]);
			ll nd=d+i*D[v];
			if (dist[v][ns]>nd) {
				dist[v][ns]=nd;
				q.emplace(nd,v,ns);
			}
		}

		// go to other city
		fore(adj[v],[u,a,b]) {
			int ns=s-a;
			ll nd=d+b;
			if (ns>=0 && dist[u][ns]>nd) {
				dist[u][ns]=nd;
				q.emplace(nd,u,ns);
			}
		}
	}

	rep(i,1,N) cout<<res[i]<<'\n';
}
