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 INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 1000007
#define INF63 1061109567
#define INF127 9187201950435737471
#define UINF 18446744073709551615
#define F first
#define S second
#define ll long long
#define N 110
using namespace std;
ll n,m,s,dp[55][N*N],c[55],d[55],sum=0;
vector<pair<ll,pair<ll,ll> > > vt[55];
priority_queue<pair<ll,pair<ll,ll> > > pq;
int main(){
	ll i,j,x,y,a,b;
	scanf("%lld%lld%lld",&n,&m,&s);
	for(i=0;i<m;i++)
	{
		scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
		sum+=a;
		x--;
		y--;
		vt[x].push_back(make_pair(y,make_pair(a,b)));
		vt[y].push_back(make_pair(x,make_pair(a,b)));
	}
	for(i=0;i<n;i++)
	{
		scanf("%lld%lld",&c[i],&d[i]);
	}
	memset(dp,127,sizeof(dp));
	dp[0][min(s,sum)]=0;
	pq.push(make_pair(0,make_pair((ll)0,min(s,sum))));
	while(!pq.empty())
	{
		x=pq.top().S.F;
		a=-pq.top().F;
		b=pq.top().S.S;
		pq.pop();
		if(dp[x][b]<a)
		{
			continue;
		}
		for(i=0;i<vt[x].size();i++)//走到另外一个地方
		{
			if(b<vt[x][i].S.F)
			{
				continue;
			}
			y=vt[x][i].F;
			if(dp[y][b-vt[x][i].S.F]>a+vt[x][i].S.S)
			{
				dp[y][b-vt[x][i].S.F]=a+vt[x][i].S.S;
				pq.push(make_pair(-dp[y][b-vt[x][i].S.F],make_pair(y,b-vt[x][i].S.F)));
			}
		}
		for(i=b+c[x];i<=sum;i+=c[x])//兑换银币
		{
			if(dp[x][i]>a+(i-b)/c[x]*d[x])
			{
				dp[x][i]=a+(i-b)/c[x]*d[x];
				pq.push(make_pair(-dp[x][i],make_pair(x,i)));
			}
		}
	}
	for(i=1;i<n;i++)
	{
		a=LINF;
		for(j=0;j<=sum;j++)
		{
			a=min(a,dp[i][j]);
		}
		printf("%lld\n",a);
	}
	return 0;
}