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 int ll
#define pb push_back
#define pii pair<int,int>
#define ff first
#define ss second
const int INF = 1e18, MAX = 2500;
vector<pair<int,pii>> graph[52];


signed main(){

	ios_base::sync_with_stdio(false);
	cin.tie(0);


	int n,m,s;
	cin>>n>>m>>s;
	vector<int> cost(n+1) , d(n+1);

	s = min(s,MAX);

	for(int i=0;i<m;i++){
		int u,v,a,b;
		cin>>u>>v>>a>>b;
		graph[u].pb({v,{a,b}});
		graph[v].pb({u,{a,b}});
	}
	for(int i=1;i<=n;i++){
		cin>>cost[i]>>d[i];
		cost[i] = min(cost[i],MAX);
	}

	vector<vector<int>> dist(n+1,vector<int> (MAX+1,INF)), visited(n+1,vector<int> (MAX+1,0));
	priority_queue<pair<int,pii>, vector<pair<int,pii>>, greater<pair<int,pii>> > pq;

	pq.push({0,{1,s}});
	dist[1][s] = 0;
	while(!pq.empty()){
		auto cur = pq.top();
		pq.pop();
		int cur_d = cur.ff, u = cur.ss.ff, cur_c = cur.ss.ss;
		if (visited[u][cur_c])
			continue;
		visited[u][cur_c] = 1;

		if (dist[u][min(cur_c+cost[u],MAX)] > cur_d + d[u]){
			dist[u][min(cur_c+cost[u],MAX)] = cur_d + d[u];
			pq.push({cur_d + d[u],{u,min(cur_c+cost[u],MAX)}});
		}


		for(auto v:graph[u]){
			if (v.ss.ff <= cur_c and dist[v.ff][cur_c - v.ss.ff] > cur_d + v.ss.ss){
				dist[v.ff][cur_c - v.ss.ff] = cur_d + v.ss.ss;
				pq.push({cur_d + v.ss.ss,{v.ff,cur_c - v.ss.ff}});
			}
		}
		
	}



	for(int j=2;j<=n;j++){
		int ans = INF;
		for(int i=0;i<MAX+1;i++)
			ans = min(ans,dist[j][i]);
		cout<<ans<<"\n";
	}

	return 0;
}