1
2 2
1 1
1 2
 1
1 3
3
1 2 3
 1
1 1
1000
1000
 1
2 2
2 2
2 2
 5
4 5
10 8 6 4
1 2 3 4 5
1 1
3
3
1 1
3
2
5 3
1000 2 2 2 3
3 1 5
5 5
1 2 3 4 5
1 2 3 4 5
 1
4 4
1 1 1 1
1 2 3 4
 1
2 3
1 1
1 2 3

YES
1 1
 YES
1 3
 YES
1 1000
 YES
1 2
 YES
1 4
YES
1 3
NO
YES
1 3
YES
1 1
 YES
1 1
 YES
1 1

def binarySearch (arr, l, r, x): 
  
    # Check base case 
    if r >= l: 
  
        mid = l + (r - l) // 2
  
        # If element is present at the middle itself 
        if arr[mid] == x: 
            return mid 
          
        # If element is smaller than mid, then it  
        # can only be present in left subarray 
        elif arr[mid] > x: 
            return binarySearch(arr, l, mid-1, x) 
  
        # Else the element can only be present  
        # in right subarray 
        else: 
            return binarySearch(arr, mid + 1, r, x) 
  
    else: 
        # Element is not present in the array 
        return -1 

t=int(input())
for x in range(t):
    n,m=input().split()
    n=int(n)
    m=int(m)
    A=list(map(int,input().split()))
    B=list(map(int,input().split()))
    A.sort()
    B.sort()
    for y in A:
        if(binarySearch(B,0,m-1,y)!=-1):
            print("YES")
            print("1",y)
            break
    else:
        print("NO")
            
