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;
using ll=long long;
vector<vector<tuple<ll,ll,ll>>> g;
vector<pair<ll,ll>> ex;
ll INF = ll(1e18)+5;
ll N,M,S;
using P =pair<ll,ll>;

vector<vector<P>> graph;


int main() {
   
    cin >> N >>M >>S;
    const ll maxCoin = N*50+1;
    graph.resize(N*maxCoin);
    for(int i=0;i<M;i++){
        ll u,v,a,b;
        cin >> u >> v >> a >>b;
        --u;--v;
        for(int j=a;j<maxCoin;j++){
            graph[u+ j*N].emplace_back((j-a)*N+v,b);
            graph[v+ j*N].emplace_back((j-a)*N+u,b);
        }
    }
    for(int i=0;i<N;i++){
        ll c,d;
        cin >> c >>d;
        for(int j=0;j<maxCoin;j++){
            graph[i+j*N].emplace_back(i+min((j+c),maxCoin-1)*N,d);
        }
    }
    priority_queue<P,vector<P>,greater<P>> pq;
    pq.push(P(0,min(maxCoin-1,S)*N));
    vector<ll> d(N*maxCoin,INF);
    d[min(maxCoin-1,S)*N] =0;
    while(!pq.empty()){
        P p = pq.top();pq.pop();
        // cerr <<"pos=" << p.second << " and" << p.second%N << endl;
        ll v =p.second;
        if(d[v] < p.first){
            continue;
        }
        for(int i=0;i<graph[v].size();i++){
            P e = graph[v][i];
            if(d[e.first] > d[v]+ e.second){
                d[e.first] = d[v] + e.second;
                pq.push(P(d[e.first],e.first));
            }
        }
    }
    for(int i=1;i<N;i++){
        ll ans = INF;
        for(int j=0;j<maxCoin;j++){
            ans = min(ans,d[i+j*N]);
        }
        cout << ans << endl;
    }
    return 0;
}