In BerSoft n programmers work, the programmer i is characterized by a skill r_i.

A programmer a can be a mentor of a programmer b if and only if the skill of the programmer a is strictly greater than the skill of the programmer b (r_a > r_b) and programmers a and b are not in a quarrel.

You are given the skills of each programmers and a list of k pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer i, find the number of programmers, for which the programmer i can be a mentor.

Input

The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k ≤ min(2 ⋅ 10^5, (n ⋅ (n - 1))/(2))) — total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers r_1, r_2, ..., r_n (1 ≤ r_i ≤ 10^{9}), where r_i equals to the skill of the i-th programmer.

Each of the following k lines contains two distinct integers x, y (1 ≤ x, y ≤ n, x ≠ y) — pair of programmers in a quarrel. The pairs are unordered, it means that if x is in a quarrel with y then y is in a quarrel with x. Guaranteed, that for each pair (x, y) there are no other pairs (x, y) and (y, x) in the input.

Output

Print n integers, the i-th number should be equal to the number of programmers, for which the i-th programmer can be a mentor. Programmers are numbered in the same order that their skills are given in the input.

Examples

Input

4 2
10 4 10 15
1 2
4 3


Output

0 0 1 2 


Input

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5


Output

5 4 0 5 3 3 9 0 2 5 

Note

In the first example, the first programmer can not be mentor of any other (because only the second programmer has a skill, lower than first programmer skill, but they are in a quarrel). The second programmer can not be mentor of any other programmer, because his skill is minimal among others. The third programmer can be a mentor of the second programmer. The fourth programmer can be a mentor of the first and of the second programmers. He can not be a mentor of the third programmer, because they are in a quarrel.
import java.util.*;
import java.lang.*;
import java.io.*;
public class Codechef
{       static PrintWriter out=new PrintWriter(System.out);static FastScanner in = new FastScanner(System.in);static class FastScanner {BufferedReader br;StringTokenizer stok;FastScanner(InputStream is) {br = new BufferedReader(new InputStreamReader(is));}
        String next() throws IOException {while (stok == null || !stok.hasMoreTokens()) {String s = br.readLine();if (s == null) {return null;}
                stok = new StringTokenizer(s);}return stok.nextToken();}
        int ni() throws IOException {return Integer.parseInt(next());}long nl() throws IOException {return Long.parseLong(next());}double nd() throws IOException {return Double.parseDouble(next());}char nc() throws IOException {return (char) (br.read());}String ns() throws IOException {return br.readLine();}
        int[] nia(int n) throws IOException{int a[] = new int[n];for (int i = 0; i < n; i++)a[i] = ni();return a;}long[] nla(int n) throws IOException {long a[] = new long[n];for (int i = 0; i < n; i++)a[i] = nl();return a;}
        double[] nda(int n)throws IOException {double a[] = new double[n];for (int i = 0; i < n; i++)a[i] = nd();return a;}int [][] imat(int n,int m) throws IOException{int mat[][]=new int[n][m];for(int i=0;i<n;i++){for(int j=0;j<m;j++)mat[i][j]=ni();}return mat;}
    }
	static long mod=Long.MAX_VALUE;
	public static void main (String[] args) throws java.lang.Exception
	{   int i,j;
	    HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
		ArrayList<Integer> arr=new ArrayList<Integer>();
        HashSet<Integer> set=new HashSet<Integer>();
        PriorityQueue<Integer> pq=new PriorityQueue<Integer>();
        int n=in.ni();
        int k=in.ni();
        int temp[]=new int[n];
        int a[]=in.nia(n);
        int b[]=new int[n];
        for(i=0;i<n;i++)
            b[i]=a[i];
        for(i=0;i<k;i++)
        {   int p=in.ni()-1;
            int q=in.ni()-1;
            if(a[p]>a[q])
                temp[p]++;
            if(a[q]>a[p])
                temp[q]++;
        }   
        sort(b);
        for(i=0;i<n;i++)
        {   int z=binarySearch(b,a[i],0,a.length-1);
            out.print(z-temp[i]+" ");
        }
        out.close();
	}
	static class pair implements Comparable<pair>{
		int x, y;
		public pair(int x, int y){this.x = x; this.y = y;}
		@Override
		public int compareTo(pair arg0) 
		{   if(x<arg0.x)    return -1;
		    else if(x==arg0.x)
		    {   if(y<arg0.y)    return -1;
		        else if(y>arg0.y)   return 1;
		        else    return 0;
		    }
		    else    return 1;
		}
	}
	static long gcd(long a,long b)
	{   if(b==0)
	        return a;
	    return gcd(b,a%b);    
	}
	static long exponent(long a,long n)
	{   long ans=1;
	    while(n!=0)
	    {   if(n%2==1)
	            ans=(ans*a)%mod;
	       a=(a*a)%mod;
	       n=n>>1;
	    }
	    return ans;
	}
	static int binarySearch(int a[], int item, int low, int high) 
    {   if (high <= low) 
            return (item > a[low])?  (low + 1): low; 
        int mid = (low + high)/2; 
        if(item > a[mid]) 
            return binarySearch(a, item, mid+1, high); 
        return binarySearch(a, item, low, mid-1); 
    } 
    static void merge(int arr[], int l, int m, int r) {   
        int n1 = m - l + 1; int n2 = r - m; int L[] = new int [n1]; int R[] = new int [n2]; 
        
        for (int i=0; i<n1; ++i) L[i] = arr[l + i]; for (int j=0; j<n2; ++j) R[j] = arr[m + 1+ j]; int i = 0, j = 0; int k = l; while (i < n1 && j < n2) {   if (L[i] <= R[j]) {   arr[k] = L[i]; i++; } else{   arr[k] = R[j]; j++; } k++; } while (i < n1){   arr[k] = L[i]; i++; k++; } while (j < n2) { arr[k] = R[j];   j++;     k++; } 
    } 
    static void Sort(int arr[], int l, int r) {if (l < r) {   int m = (l+r)/2; Sort(arr, l, m); Sort(arr , m+1, r); merge(arr, l, m, r); } } 
    
    static void sort(int a[])
    {Sort(a,0,a.length-1);}
}