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 endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const int N=200000+5;
const ll oo=1e18+5;
const ll mod=998244353;

vector<pair<int,int>> v[55];
ll dp[55][2505];

struct node{
	ll val;
	int id,sum;
	node(ll val,int id,int sum): val(val),id(id),sum(sum) {}
};

void min_self(int& a,int b){
	if(a>b) a=b;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	const int maxx=2500;
	int n,m,s;
	cin>>n>>m>>s;
	vector<int> a(105),b(105),c(55),d(55);
	for(int i=0;i<m;++i){
		int x,y;
		cin>>x>>y>>a[i]>>b[i];
		--x;
		--y;
		v[x].push_back({y,i});
		v[y].push_back({x,i});
	}
	for(int i=0;i<n;++i) cin>>c[i]>>d[i];
	min_self(s,maxx);
	for(int i=0;i<n;++i){
		for(int j=0;j<=maxx;++j) dp[i][j]=oo;
	}
	queue<node> Q;
	dp[0][s]=0;
	Q.push({0,0,s});
	while(!Q.empty()){
		node ele=Q.front();
		Q.pop();
		int idx=ele.id;
		int coin=ele.sum;
		if(ele.val>dp[idx][coin]) continue;
		int n_coin=min(maxx,coin+c[idx]);
		if(dp[idx][n_coin]>dp[idx][coin]+d[idx]){
			dp[idx][n_coin]=dp[idx][coin]+d[idx];
			Q.push({dp[idx][n_coin],idx,n_coin});
		}
		for(pair<int,int> x:v[idx]){
			int xx=x.first;
			int idd=x.second;
			if(coin<a[idd]) continue;
			if(dp[xx][coin-a[idd]]>dp[idx][coin]+b[idd]){
				dp[xx][coin-a[idd]]=dp[idx][coin]+b[idd];
				Q.push({dp[xx][coin-a[idd]],xx,coin-a[idd]});
			}
		}
	}
	for(int i=1;i<n;++i){
		ll ans=oo;
		for(int coin=0;coin<=maxx;++coin){
			ans=min(ans,dp[i][coin]);
		}
		cout<<ans<<endl;
	}
	return 0;
}
