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;
struct nobe
{
    int v,a,b;
};
vector<nobe> gg[55];
int c[55],d[55];
long long f[55][3030];
int main()
{
    int n,m,s,i=1,j=0;
    scanf("%d%d%d",&n,&m,&s);
	s=min(s,2500);
    while(i<=m)
	{
		int u,v,a,b;
        scanf("%d%d%d%d",&u,&v,&a,&b);
        gg[u].push_back((nobe){v,a,b});
        gg[v].push_back((nobe){u,a,b});
        ++i;
    }
    i=1;
    while(i<=n)
    {
    	scanf("%d%d",&c[i],&d[i]);
    	++i;
    }
    memset(f,63,sizeof(f));
    f[1][s]=0;
    typedef tuple<long long,int,int> tt;
    priority_queue<tt,vector<tt>,greater<tt> > qu;
    qu.push(tt(0,1,s));
    while(!qu.empty())
	{
        long long t=get<0>(qu.top());
        int u=get<1>(qu.top()),w=get<2>(qu.top());
        qu.pop();
        if(f[u][w]>t) continue;
        i=0;
        while(i<gg[u].size())
		{
            int v=gg[u][i].v,a=gg[u][i].a,b=gg[u][i].b;
            if((w>=a)&&(f[v][w-a]>t+b))
			{
                f[v][w-a]=t+b;
                qu.push(tt(t+b,v,w-a));
            }
            ++i;
        }
        if(f[u][min(w+c[u],2500)]>t+d[u])
		{
            f[u][min(w+c[u],2500)]=t+d[u];
            qu.push(tt(t+d[u],u,min(w+c[u],2500)));
        }
    }
    i=2;
    while(i<=n)
	{
        long long ans=999999999999999999ll;
        j=0;
        while(j<=2500)
		{
            ans=min(ans,f[i][j]);
            ++j;
        }
        printf("%lld\n",ans);
		++i;
    }
    return 0;
}
/*
3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5
*/