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(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
typedef long long ll;
typedef pair<int,int> pii;

struct Edge{
    int to,mai,t;
    Edge(){}
    Edge(int to,int mai,int t){
        this->to = to;
        this->mai = mai;
        this->t = t;
    }
};

typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

typedef tuple<ll,ll,ll> T;
const ll INF = 1e+15;
ll dp[50][2501];
ll c[50],d[50];
int main(){
    int n,m,s;
    cin>>n>>m>>s;

    Graph g(n);
    rep(i,m){
        int u,v,a,b;
        cin>>u>>v>>a>>b;
        u--;v--;
        g[u].push_back(Edge(v,a,b));
        g[v].push_back(Edge(u,a,b));
    }
    rep(i,n) cin>>c[i]>>d[i];


    rep(i,n) rep(j,2501) dp[i][j]=INF;

    s = min(s,2500);
    priority_queue<T,vector<T>,greater<T>> que;
    que.push(T(0,0,s));

    while(!que.empty()){
        T now = que.top();que.pop();

        ll t = get<0>(now);
        ll pos =get<1>(now);
        ll mai = get<2>(now);
        if(dp[pos][mai]<=t) continue;
        dp[pos][mai]=t;
        rep(i,g[pos].size()){
            Edge e = g[pos][i];
            if(mai>=e.mai){
                que.push(T(t+e.t,e.to,mai-e.mai));
            } 
        }
        que.push(T(t+d[pos],pos,min((ll)2500,mai+c[pos])));
    }
    
    for(int i = 1;i<n;i++){
        ll ans = INF;
        rep(j,2501){
            ans = min(ans,dp[i][j]);
        }
        cout<<ans<<endl;
    }
    return 0;

}