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;
#define rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll n,m,S;
vec U(100),V(100),A(100),B(100),C(50),D(50);
struct edge{ll to,cost,money;};
vector<edge> G[200001];
ll d[51][2501],INF=1e18;
int main(){
    cin>>n>>m>>S;
    rep(i,m) cin>>U[i]>>V[i]>>A[i]>>B[i],G[U[i]].push_back({V[i],B[i],A[i]}),G[V[i]].push_back({U[i],B[i],A[i]});
    rep(i,n) cin>>C[i+1]>>D[i+1];
    priority_queue<pair<ll,P>,vector<pair<ll,P>>,greater<pair<ll,P>>> que;
    rep(i,n) fill(d[i+1],d[i+1]+2501,INF);
    if(S>2500) S=2500;
    d[1][S]=0;
    que.push(make_pair(0,P(1,S)));
    while(!que.empty()){
        pair<ll,P> p=que.top();que.pop();
        ll v=p.second.first,money=p.second.second;
        if(d[v][money]<p.first) continue;
        rep(i,G[v].size()){
            edge e=G[v][i];
            if(money>=e.money&&d[e.to][money-e.money]>d[v][money]+e.cost){
                d[e.to][money-e.money]=d[v][money]+e.cost,que.push(make_pair(d[e.to][money-e.money],P(e.to,money-e.money)));
            }
        }
        for(ll i=1;true;i++){
            if(money+C[v]*i>2500) break;
            if(d[v][money+C[v]*i]>d[v][money]+D[v]*i) d[v][money+C[v]*i]=d[v][money]+D[v]*i,que.push(make_pair(d[v][money+C[v]*i],P(v,money+C[v]*i)));
        }
    }
    rep(i,n-1){
        ll x=INF;
        rep(j,2501) x=min(x,d[i+2][j]);
        cout<<x<<"\n";
    }
}