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 int long long
#define f first
#define s second
using namespace std;
const int X=2509,N=55,M=105,inf=1e18;
int n,m,s;
int a[M],b[M],c[N],d[N],t[N*X];
vector <pair <int,int> > g[N];
int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    cin >> n >> m >> s;
    s=min(X-1,s);
    for(int i=0;i<m;i++){
        int u,v;
        cin >> u >> v >> a[i] >> b[i];
        g[u].push_back({v,i});
        g[v].push_back({u,i});
    }
    for(int i=1;i<=n;i++){
        cin >> c[i] >> d[i];
        c[i]=min(c[i],X-1);
    }
    fill(t,t+N*X,inf);
    t[1*X+s]=0;
    set <pair <int,int> > q;
    q.insert({0,1*X+s});
    while(!q.empty()){
        auto k=q.begin()->s;
        q.erase(q.begin());
        int v=k/X,cm=k%X;
        if(cm<X-1){
            int nm=cm+c[v];
            if(nm>=X)nm=X-1;
            int pos=v*X+nm;
            if(t[k]+d[v]<t[pos]){
                q.erase({t[pos],pos});
                t[pos]=t[k]+d[v];
                q.insert({t[pos],pos});
            }
        }
        for(auto ed:g[v]){
            int u=ed.f,i=ed.s,nm=cm-a[i];
            if(nm<0)continue;
            int pos=u*X+nm;
            if(t[k]+b[i]<t[pos]){
                q.erase({t[pos],pos});
                t[pos]=t[k]+b[i];
                q.insert({t[pos],pos});
            }
        }
    }
    for(int i=2;i<=n;i++){
        int ans=inf;
        for(int pos=i*X;pos<(i+1)*X;pos++)
            ans=min(ans,t[pos]);
        cout << ans << '\n';
    }
}