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;
long long n,m,dp[53][2503],s;
vector<pair<long long,pair<long long,long long> > >v[53];
int main(){
	scanf("%lld%lld%lld",&n,&m,&s);
	s=min(s,n*50);
	while(m--){
		long long x,y,a,b;scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
		v[x-1].push_back(make_pair(y-1,make_pair(-a,b)));
		v[y-1].push_back(make_pair(x-1,make_pair(-a,b)));
	}
	memset(dp,31,sizeof(dp));
	for(long long i=0;i<n;i++)
	{
		long long c,d;cin>>c>>d;
		v[i].push_back(make_pair(i,make_pair(c,d)));
	}
	priority_queue<pair<long long,pair<long long,long long> > >Q;
	Q.push(make_pair(0,make_pair(0,s)));
	dp[0][s]=0;
	while(Q.size()){
		long long tm=Q.top().first,ct=Q.top().second.first,cn=Q.top().second.second;Q.pop();
		for(long long i=0;i<v[ct].size();i++){
			long long nct=v[ct][i].first,ntm=tm-v[ct][i].second.second,ncoi=v[ct][i].second.first+cn;
			if(ncoi<0)continue;
			if(ncoi>n*50)continue;
			ncoi=min(ncoi,n*50);
			if(dp[nct][ncoi]>-ntm){
				dp[nct][ncoi]=-ntm; 
				Q.push(make_pair(ntm,make_pair(nct,ncoi)));
			}
		}
	}
	for(long long i=1;i<n;i++){
		long long ans=LONG_LONG_MAX;
		for(long long j=0;j<=n*50;j++)ans=min(ans,dp[i][j]);
		cout<<ans<<endl;
	}
}