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 int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 4223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

int amax=0;
struct edge { ll to, cost, time; };
vector<edge> G[200];
vector<P> cd(200);
VI d(600000,LINF);

void dijkstra(int s) {
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P> > que;
	que.push(P(0, s));
	while (!que.empty()) {
		P p = que.top();
		que.pop();
		int v = p.second/amax;
		if (d[p.second] < p.first) continue;
		REP(i,G[v].size()) {
			edge e = G[v][i];
			bool ok=1;
			int have = p.second%amax;
			for(int j=0;ok;j++){
				if(have>=amax){
					ok=0;
					have=amax-1;
				}
				if(have>=e.cost){
					if(d[e.to*amax+have-e.cost]>d[p.second]+e.time+(cd[v].second*j)){
						d[e.to*amax+have-e.cost]=d[p.second]+e.time+(cd[v].second*j);
						que.push(P(d[e.to*amax+have-e.cost], e.to*amax+have-e.cost));
					}
				}
				have+=cd[v].first;
			}
		}
	}
}

int main() {
	int n, m, s; 
	cin >> n >> m >> s;
	REP(i,m){
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		u--; v--;
		G[u].push_back({v,a,b});
		G[v].push_back({u,a,b});
		amax=max(amax,a*(n-1));
	}
	REP(i,n){
		cin >> cd[i].first >> cd[i].second;
	}
	amax++;
	dijkstra(min(s,amax-1));
	for(int i=1;i<n;i++){
		ll ans=LINF;
		REP(j,amax){
			ans=min(ans,d[i*amax+j]);
		}
		cout << ans << endl;
	}
	return 0;
}
