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 fr(i,n) for(int i = 0; i<n; i++)
#define sz(v) (int)(v.size())
#define prin(a) cout << #a << " = " << a << endl
#define all(v) (v).begin(),(v).end()
typedef long long ll;

#define get_money iwfebvosbi

const int N = 50, M = 5e3;
vector<tuple<ll,ll,ll>> g[N+10];

ll n, m, s;
bool vis[N+10][M+10];
ll t_dist[N+10][M+10];
ll get_money[N], time_get_money[N];

void dij(){
	s = min<ll>(s,M);
	set<tuple<ll,ll,ll>> st;
	st.emplace(0,0,s);
	
	while(!st.empty()){
		ll curt, no, cur_money;
		tie(curt,no,cur_money) = *(st.begin());
		st.erase(st.begin());
		if(vis[no][cur_money]) continue;
		vis[no][cur_money] = 1;
		t_dist[no][cur_money] = curt;
		
		for(auto &tup : g[no]){
			ll it,cost,temp;
			tie(it,cost,temp) = tup;
			if(cur_money>=cost) st.emplace(curt+temp,it,cur_money-cost);
		}	
		if(cur_money+get_money[no]<=M){
			st.emplace(curt+time_get_money[no],no,cur_money+get_money[no]);
		}
	}
}

int main(){
	cin >> n >> m >> s;
	
	fr(i,m){
		ll u, v, cost, temp;
		scanf("%lld%lld%lld%lld", &u, &v, &cost, &temp);
		u--,v--;
		g[u].emplace_back(v,cost,temp);
		g[v].emplace_back(u,cost,temp);
	}
	fr(i,n){
		scanf("%lld%lld", get_money+i, time_get_money+i);
	}
	
	dij();
	
	for(int i = 1; i<n; i++){
		ll mint = LLONG_MAX;
		fr(din,M){
			if(vis[i][din]) mint = min(mint,t_dist[i][din]);
		}
		printf("%lld\n", mint);
	}
}