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;
typedef long long ll;
typedef double db;
ll ans[55],c[55],d[55],f[55][10005];
struct E{ll to,a,b;};
struct P{ll t,pos,s;};
bool operator <(const P &a,const P &b){return a.t>b.t;}
vector<E> edge[55];
priority_queue<P> q;
int main(){
    ll n,m,k,sum=0;
    cin>>n>>m>>k;
    for(ll i=1;i<=m;i++){
        ll u,v,x,y;
        cin>>u>>v>>x>>y;
        edge[u].push_back({v,x,y});
        edge[v].push_back({u,x,y});
    }
    for(ll i=1;i<=n;i++){
        cin>>c[i]>>d[i];
        c[i]=min(c[i],(ll)5000);
    }
    k=min(k,(ll)5000);
    q.push({1,1,k});
    while(!q.empty()&&sum<n){
        ll x=q.top().t,p=q.top().pos,num=q.top().s;q.pop();
        if(f[p][num]) continue;
        f[p][num]=x;
        if(!ans[p]) ans[p]=x,sum++;
        for(E i:edge[p]){
            if(i.a<=num&&!f[i.to][num-i.a]) q.push({x+i.b,i.to,num-i.a});
            if(num+c[p]<=10000) q.push({x+d[p],p,num+c[p]});
        }
    }
    for(int i=2;i<=n;i++){
        cout<<ans[i]-1<<endl;
    }
}