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 <stdio.h>
#include <vector>
#include <queue>
struct str{
	int x0;
	int cost;
	int time;
};
std::vector<str> V[60];
struct str2{
	int x0;
	int coin;
	long long int dist;
};
bool operator<(str2 a, str2 b)
{
	return a.dist>b.dist;
}
std::priority_queue<str2> Q;
std::pair<int,int> x[60];
long long int check[60][3000];

long long int MAX = 1;
int main()
{
	for(int i=1;i<=15;i++) MAX*=10;
	
	int a,b,c;
	scanf("%d%d%d",&a,&b,&c);
	for(int i=1;i<=b;i++)
	{
		int d,e,f,g;
		scanf("%d%d%d%d",&d,&e,&f,&g);
		V[d].push_back({e,f,g});
		V[e].push_back({d,f,g});
	}
	for(int i=1;i<=a;i++)
	{
		int d,e;
		scanf("%d%d",&d,&e);
		x[i]=std::make_pair(d,e);
	}
	
	if(c>2500) c = 2500;
	Q.push({1,c,1});
	while(!Q.empty())
	{
		int x0 = Q.top().x0;
		int coin = Q.top().coin;
		long long int dist = Q.top().dist;
		Q.pop();
		
		if(check[x0][coin]) continue;
		check[x0][coin] = dist;
		
		if(coin+x[x0].first <= 2500) Q.push({x0,coin+x[x0].first,dist+x[x0].second});
		for(int i=0;i<V[x0].size();i++) if(V[x0][i].cost<=coin) Q.push({V[x0][i].x0,coin-V[x0][i].cost,dist+V[x0][i].time});
	}
	
	for(int i=2;i<=a;i++)
	{
		long long int ans = MAX;
		for(int j=0;j<=2500;j++)
		{
			if(check[i][j]>0) ans = ans<check[i][j]?ans:check[i][j];
		}
		printf("%lld\n",ans-1);
	}
}