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;
typedef long long ll;
#define REP(i,n) for(ll i = 0; i<(ll)(n); i++)
#define REPS(i,n) for (ll i = 1; i<=(ll)(n); i++)
#define REPD(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define FORD(i,a,b) for(ll i=(a);i>=(b);i--)
#define pb push_back
#define MOD 1000000007
#define MOD2 998244353
#define PI 3.141592653
#define INF 100000000000000

int main(){
	vector<vector<vector<ll>>> conn(50);
	vector<vector<ll>> exch(50,vector<ll>(2));
	vector<ll> ansl(250000,INF);
	priority_queue<vector<ll>, vector<vector<ll>>, greater<vector<ll>>> q;
	ll n, m, s, u, v, x, y, a, b, c, d; cin >> n >> m >> s;
	ll slim = 4999;
	s = min(s,slim);
	REP(i,m){
		cin >> u >> v >> a >> b; u--; v--;
		conn[u].pb({v,a,b});
		conn[v].pb({u,a,b});
	}
	REP(i,n) cin >> exch[i][0] >> exch[i][1];
	q.push({0,s});
	while(q.size()){
		ll nmin = q.top().at(0);
		ll nnod = q.top().at(1);
		//cout << nmin << " " << nnod << "check" << endl;
		ll nowat = nnod/5000;
		x = min(nnod+exch[nowat][0],5000*nowat+4999);
		y = nmin+exch[nowat][1];
		if (y<ansl[x]){
			ansl[x]=y;
			q.push({y,x});
			//cout << y << " " << x << endl;
		}
		REP(i,conn[nowat].size()){
			if (nnod%5000>=conn[nowat][i][1]){
				x = nnod-conn[nowat][i][1]+5000*(conn[nowat][i][0]-nowat);
				y = nmin+conn[nowat][i][2];
				if (y<ansl[x]){
					ansl[x]=y;
					q.push({y,x});
				}
			}
		}
		q.pop();
	}
	FOR(i,1,n-1){
		ll ans = INF;
		REP(j,5000){
			ans = min(ans,ansl[i*5000+j]);
		}
		cout << ans << endl;
	}
}