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>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;
typedef pair<int,int>P;
typedef long long ll;

struct st{
	ll a,b,c;
};

bool operator<(st a,st b){
	return a.c>b.c;
}
vector<st>E[60];
ll dp[60][6000];
ll c[60],d[60];

int main(){
	int n,m,s;cin>>n>>m>>s;s=min(s,5500);
	rep(i,m){
		int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);a--;b--;
		E[a].push_back({b,c,d});
		E[b].push_back({a,c,d});
	}
	rep(i,n)scanf("%lld%lld",&c[i],&d[i]);
	memset(dp,0x3f,sizeof(dp));
	priority_queue<st>que;
	dp[0][s]=0;
	que.push({0,s,0});
	while(!que.empty()){
		st p=que.top();que.pop();
		if(dp[p.a][p.b]!=p.c)continue;
		if(p.b+c[p.a]<5500){
			if(dp[p.a][p.b+c[p.a]]>p.c+d[p.a]){
				dp[p.a][p.b+c[p.a]]=p.c+d[p.a];
				que.push({p.a,p.b+c[p.a],p.c+d[p.a]});
			}
		}
		for(auto u:E[p.a]){
			if(p.b-u.b>=0){
				if(dp[u.a][p.b-u.b]>p.c+u.c){
					dp[u.a][p.b-u.b]=p.c+u.c;
					que.push({u.a,p.b-u.b,p.c+u.c});
				}
			}
		}
	}
	for(int i=1;i<n;i++){
		cout<<*min_element(dp[i],dp[i]+5500)<<endl;
	}
}