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;
constexpr ll mod=1000000007;

struct edge{
	int to;
	int cost;
	ll dist;
};


int N,M,S;
ll C[50],D[50],T[50][3001];
vector<edge>E[50];

int main(){
	cin>>N>>M>>S;
	while(M--){
		int u,v,a,b;
		cin>>u>>v>>a>>b;
		u--;
		v--;
		E[u].push_back({v,a,b});
		E[v].push_back({u,a,b});
	}
	for(int i=0;i<N;i++){
		cin>>C[i]>>D[i];
	}
	for(int i=0;i<50;i++){
		for(int j=0;j<=3000;j++){
			T[i][j]=-1;
		}
	}
	priority_queue<pair<ll,pair<int,int>>,vector<pair<ll,pair<int,int>>>,greater<pair<ll,pair<int,int>>>>pq;
	pq.push({0,{0,S}});
	while(!pq.empty()){
		ll t=pq.top().first;
		int x=pq.top().second.first;
		int m=min(3000,pq.top().second.second);
		pq.pop();
		if(T[x][m]==-1){
			T[x][m]=t;
			pq.push({t+D[x],{x,m+C[x]}});
			for(edge e:E[x]){
				if(e.cost<=m){
					pq.push({t+e.dist,{e.to,m-e.cost}});
				}
			}
		}
	}
	for(int i=1;i<N;i++){
		ll res=0xE869120E869120;
		for(int j=0;j<=3000;j++){
			if(T[i][j]!=-1){
				res=min(res,T[i][j]);
			}
		}
		cout<<res<<endl;
	}
}
