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;
using ll = long long;
using tp = tuple<ll,ll,ll>;
ll n,m,s;
ll a,b; 
ll c[55],d[55];       //存获得的银币和消耗的时间 
ll dp[55][2505];            //到达i时还有j银币的最小时间 
vector<tp> G[55];          //目标点 a b 
tp info;              //时间 目标点 当前的钱 
const ll inf = 0x3f3f3f3f3f3f3f3f;
int main(){
	cin>>n>>m>>s;
	s = min(s,2504LL);
	for(ll i=0;i<m;i++){
		ll ta,tb,tc,td;
		cin>>ta>>tb>>tc>>td;
		G[ta].push_back(tp(tb,tc,td));
		G[tb].push_back(tp(ta,tc,td));	 
	}
	for(ll i=1;i<=n;i++){
		ll tc,td;
		cin>>tc>>td;
		c[i] = tc;
		d[i] = td;
	}
	fill(dp[0] , dp[0] + 54*2504 , inf);
	dp[1][s] = 0;
	priority_queue<tp,vector<tp>,greater<tp>> pq;
	pq.push(tp(0,1,s));
	while(!pq.empty()){
		info = pq.top();
		pq.pop();
		ll t = get<0>(info) , u = get<1>(info) , w = get<2>(info);
		if(dp[u][w] > t) continue;
		for(auto i:G[u]){
			ll v = get<0>(i) , aa = get<1>(i) , bb = get<2>(i);
			if(w>=aa and dp[v][w-aa] > t + bb){
				dp[v][w-aa] = t + bb;
				pq.push(tp(t+bb , v , w-aa));
			}
		}
		if(dp[u][min(w + c[u] , 2504LL)] > t + d[u]){
			dp[u][min(w + c[u] , 2504LL)] = t + d[u];
			pq.push(tp(t + d[u] , u , min(w + c[u] , 2504LL)));
		}
		
	
	}	
	
	for(ll i=2;i<=n;i++){
			ll ans = inf;
			for(ll j=0;j<2504;j++){
				ans = min(ans , dp[i][j]);
			}
			cout<<ans<<endl;
		}
	
	return 0;
}





