Problem: BigData Inc. is a corporation that has n data centers indexed from 1 to n that are located all over the world. These data centers provide storage for client data (you can figure out that client data is really big!).

Main feature of services offered by BigData Inc. is the access availability guarantee even under the circumstances of any data center having an outage. Such a guarantee is ensured by using the two-way replication. Two-way replication is such an approach for data storage that any piece of data is represented by two identical copies that are stored in two different data centers.

For each of m company clients, let us denote indices of two different data centers storing this client data as ci, 1 and ci, 2.

In order to keep data centers operational and safe, the software running on data center computers is being updated regularly. Release cycle of BigData Inc. is one day meaning that the new version of software is being deployed to the data center computers each day.

Data center software update is a non-trivial long process, that is why there is a special hour-long time frame that is dedicated for data center maintenance. During the maintenance period, data center computers are installing software updates, and thus they may be unavailable. Consider the day to be exactly h hours long. For each data center there is an integer uj (0 ≤ uj ≤ h - 1) defining the index of an hour of day, such that during this hour data center j is unavailable due to maintenance.

Summing up everything above, the condition uci, 1 ≠ uci, 2 should hold for each client, or otherwise his data may be unaccessible while data centers that store it are under maintenance.

Due to occasional timezone change in different cities all over the world, the maintenance time in some of the data centers may change by one hour sometimes. Company should be prepared for such situation, that is why they decided to conduct an experiment, choosing some non-empty subset of data centers, and shifting the maintenance time for them by an hour later (i.e. if uj = h - 1, then the new maintenance hour would become 0, otherwise it would become uj + 1). Nonetheless, such an experiment should not break the accessibility guarantees, meaning that data of any client should be still available during any hour of a day after the data center maintenance times are changed.

Such an experiment would provide useful insights, but changing update time is quite an expensive procedure, that is why the company asked you to find out the minimum number of data centers that have to be included in an experiment in order to keep the data accessibility guarantees.

Input

The first line of input contains three integers n, m and h (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000, 2 ≤ h ≤ 100 000), the number of company data centers, number of clients and the day length of day measured in hours. 

The second line of input contains n integers u1, u2, ..., un (0 ≤ uj < h), j-th of these numbers is an index of a maintenance hour for data center j. 

Each of the next m lines contains two integers ci, 1 and ci, 2 (1 ≤ ci, 1, ci, 2 ≤ n, ci, 1 ≠ ci, 2), defining the data center indices containing the data of client i.

It is guaranteed that the given maintenance schedule allows each client to access at least one copy of his data at any moment of day.

Output

In the first line print the minimum possible number of data centers k (1 ≤ k ≤ n) that have to be included in an experiment in order to keep the data available for any client.

In the second line print k distinct integers x1, x2, ..., xk (1 ≤ xi ≤ n), the indices of data centers whose maintenance time will be shifted by one hour later. Data center indices may be printed in any order.

If there are several possible answers, it is allowed to print any of them. It is guaranteed that at there is at least one valid choice of data centers.

Examples

Input

3 3 5
4 4 0
1 3
3 2
3 1


Output

1
3 

Input

4 5 4
2 1 0 3
4 3
3 2
1 2
1 4
1 3


Output

4
1 2 3 4 

Note

Consider the first sample test. The given answer is the only way to conduct an experiment involving the only data center. In such a scenario the third data center has a maintenance during the hour 1, and no two data centers storing the information of the same client have maintenance at the same hour.

On the other hand, for example, if we shift the maintenance time on hour later for the first data center, then the data of clients 1 and 3 will be unavailable during the hour 0.

Solution (in JAVA):

Solution:
/* Author: Ronak Agarwal, reader part not written by me*/
import java.io.* ; import java.util.* ;
import static java.lang.Math.min ; import static java.lang.Math.max ;
import static java.lang.Math.abs ; import static java.lang.Math.log ;
import static java.lang.Math.pow ; import static java.lang.Math.sqrt ;
/* Thread is created here to increase the stack size of the java code so that recursive dfs can be performed */
public class Contest{
	public static void main(String[] args) throws IOException{
		new Thread(null,new Runnable(){	public void run(){ exec_Code() ;} },"Solver",1l<<27).start() ;
	}
	static void exec_Code(){
		try{
			Solver Machine = new Solver() ;
			Machine.Solve() ; Machine.Finish() ;}
		catch(Exception e){ e.printStackTrace() ; print_and_exit() ;}
		catch(Error e){ e.printStackTrace() ; print_and_exit() ; }
	}
	static void print_and_exit(){ System.out.flush() ; System.exit(-1) ;}
}
/* Implementation of the Reader class */
class Reader { 
	private InputStream mIs;
	private byte[] buf = new byte[1024];
	private int curChar,numChars;
	public Reader() { this(System.in); } 
	public Reader(InputStream is) { mIs = is;} 
	public int read() {
		if (numChars == -1) throw new InputMismatchException();    
		if (curChar >= numChars) {
			curChar = 0;
			try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}
			if (numChars <= 0) return -1;            
		}
		return buf[curChar++];
	} 
	public String nextLine(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString() ;
	} 
	public String s(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		}while (!isSpaceChar(c));
		return res.toString();
	} 
	public long l(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		long res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read();
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int i(){
		int c = read() ;
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		int res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read() ;
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
	public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
}   
/* All the useful functions,constants,renamings are here*/
class Template{
	/* Constants Section */
	final int INT_MIN = Integer.MIN_VALUE ; final int INT_MAX = Integer.MAX_VALUE ;
	final long LONG_MIN = Long.MIN_VALUE ; final long LONG_MAX = Long.MAX_VALUE ;
	static long MOD = 1000000007 ;
	static Reader ip = new Reader() ;	
	static PrintWriter op = new PrintWriter(System.out) ;
	/* Methods for writing */
	static void p(Object o){ op.print(o) ; }
	static void pln(Object o){ op.println(o) ;}
	static void Finish(){ op.flush(); op.close(); }
	/* Implementation of operations modulo MOD */
	static long inv(long a){ return powM(a,MOD-2) ; }
	static long m(long a,long b){ return (a*b)%MOD ; }
	static long d(long a,long b){ return (a*inv(b))%MOD ; }
	static long powM(long a,long b){
		if(b<0) return powM(inv(a),-b) ;
		long ans=1 ;
		for( ; b!=0 ; a=(a*a)%MOD , b/=2) if((b&1)==1) ans = (ans*a)%MOD ;
		return ans ;
	}
	/* Renaming of some generic utility classes */
	final static class mylist extends ArrayList<Integer>{}
	final static class myset extends TreeSet<Integer>{}
	final static class mystack extends Stack<Integer>{}
	final static class mymap extends TreeMap<Integer,Integer>{}
}
/* Implementation of the pair class, useful for every other problem */
class pair implements Comparable<pair>{
	int x ; int y ;
	pair(int _x,int _y){ x=_x ; y=_y ;} 
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
	}
}
/* Main code starts here */
class Solver extends Template{
	final int MAXN = 100005 ;
	mylist out[] = new mylist[MAXN] ;
	mylist in[] = new mylist[MAXN] ;
	int u[] = new int[MAXN] ;
	int topo[] = new int[MAXN] ;
	boolean vis[] = new boolean[MAXN] ;
	int idx ;
	int mrk=0 ;
	int comp[] = new int[MAXN] ;
	int cnt[] = new int[MAXN] ;
	int odeg[] = new int[MAXN] ;
	void dfs(int nd){
		vis[nd]=true ;
		for(int ng : out[nd]) if(!vis[ng]) dfs(ng) ;
		topo[idx--] = nd ;
	}
	void dfs2(int nd){
		comp[nd]=mrk ; cnt[mrk]++ ;
		for(int ng : in[nd]) if(comp[ng]==0) dfs2(ng) ;
	}
	public void Solve() throws IOException{
		int n = ip.i() ; int m = ip.i() ; int h = ip.i() ;
		for(int i=1 ; i<=n ; i++) u[i] = ip.i() ;
		for(int i=1 ; i<=n ; i++) out[i] = new mylist() ;
		for(int i=1 ; i<=n ; i++) in[i] = new mylist() ;
		for(int i=1 ; i<=m ; i++){
			int x = ip.i() ; int y = ip.i() ;
			int ux = u[x] ; int uy = u[y] ;
			if(((uy-ux)+h)%h==1){
				out[x].add(y) ;
				in[y].add(x) ;
			}
			if(((ux-uy)+h)%h==1){ 
				out[y].add(x) ;
				in[x].add(y) ;
			}
		}
		idx=n-1 ;
		for(int i=1 ; i<=n ; i++) if(!vis[i]) dfs(i) ;
		for(int i=0 ; i<n ; i++){
			int v = topo[i] ;
			if(comp[v]==0){
				++mrk ;
				dfs2(v) ;
			}
		}
		// for(int i=0 ; i<n ; i++) p(topo[i]+" ") ;
		// pln("") ;
		// for(int i=1 ; i<=n ; i++) p(comp[i]+" ") ;
		// pln("") ;
		for(int i=1 ; i<=n ; i++){
			int j = comp[i] ;
			for(int chd : out[i])
				if(comp[chd]!=j)
					odeg[j]++ ;
		}
		int res = -1 ; int mn = INT_MAX ;
		for(int i=1 ; i<=mrk ; i++) if(odeg[i]==0 && cnt[i]<mn){
			res = i ; mn = cnt[i] ;
		}
		pln(mn) ;
		for(int i=1 ; i<=n ; i++) if(comp[i]==res) p(i+" ") ;
		pln("") ;
	}
}