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;
const int N = 2e5 + 10;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
int n,m,S;
struct uzi{
	int t,a,b;
};
struct wzh{
	int a,b;
	LL d;
	bool operator < (const wzh &t) const {
		return d>t.d;
	}
};
int c[N],d[N];
LL dis[55][55];
vector<uzi>v[N];
LL dp[55][51*51];
void dfs(int s){
	priority_queue<wzh>q;
  int mx=n*50;
	memset(dp,0x3f3f3f3f,sizeof dp);
  S=min(S,mx);
	dp[1][S]=0;
	q.push({1,S,0ll});
	while(!q.empty()){
		auto x=q.top();
		q.pop();
		for(auto k:v[x.a]){
			int less=x.b-k.a;
			LL dis=x.d+k.b,up=0;
			int fl=0;
			while(less<=mx){
				if(less>=0){
					if(dp[k.t][less]>dis+up){
						dp[k.t][less]=dis+up;
						q.push({k.t,less,dp[k.t][less]});
					}
        }
				up+=d[x.a];
				less+=c[x.a];
				if(less>mx&&!fl){
					less=mx;
					fl=1;
				}
			}
		}
	}
}
int main(){
  ios::sync_with_stdio(false);
  cin>>n>>m>>S;
  for(int i=1;i<=m;i++){
  	int s,t,a,b;
  	cin>>s>>t>>a>>b;
  	v[s].pb({t,a,b});
  	v[t].pb({s,a,b});
  }
  for(int i=1;i<=n;i++){
  	cin>>c[i]>>d[i];
  }
  dfs(1);
  for(int i=2;i<=n;i++){
  	LL ans=1e18;
  	for(int j=0;j<=50*50;j++){
  		ans=min(ans,dp[i][j]);
  	}
  	cout<<ans<<'\n';
  }
 	return 0;
}