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
#pragma GCC optimize("Ofast")
#include<stdio.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
struct Edge{
	long long to,cost,dis;//used in state is cur_v,cur_money,distance with node 1
};
vector<Edge> g[55];
long long c[55],d[55],dp[55][2550],maxv=2500;
struct cmp{
	bool operator()(const Edge a,const Edge b)const{
		return a.dis>b.dis;
	}
};
priority_queue<Edge,vector<Edge>,cmp> pq;
int main(){
	long long i,j,n,m,s,u,v,a,b;
	scanf("%lld%lld%lld",&n,&m,&s);
	for(i=0;i<m;i++){
		scanf("%lld%lld%lld%lld",&u,&v,&a,&b);
		g[u].emplace_back((Edge){v,a,b});
		g[v].emplace_back((Edge){u,a,b});
	}
	for(i=1;i<=n;i++)	
		scanf("%lld%lld",&c[i],&d[i]);
	for(i=0;i<55;i++)
		for(j=0;j<2550;j++)
			dp[i][j]=100000000000000000;
	s=min(s,maxv);
	pq.push({1,s,0});
	dp[1][s]=0;
	while(pq.size()){
		auto x=pq.top();
		pq.pop();
		if(dp[x.to][min(x.cost+c[x.to],maxv)]>x.dis+d[x.to]){
			dp[x.to][min(x.cost+c[x.to],maxv)]=x.dis+d[x.to];
			pq.push({x.to,min(x.cost+c[x.to],maxv),dp[x.to][min(x.cost+c[x.to],maxv)]});
		}
		for(auto y:g[x.to]){
			if(x.cost<y.cost)
				continue;
			if(dp[y.to][x.cost-y.cost]>x.dis+y.dis){
				dp[y.to][x.cost-y.cost]=x.dis+y.dis;
				pq.push({y.to,x.cost-y.cost,dp[y.to][x.cost-y.cost]});
			}
		}
	}
	for(i=2;i<=n;i++)
		printf("%lld\n",*min_element(dp[i],dp[i]+maxv));
}