You might have remembered Theatre square from the [problem 1A](https://codeforces.com/problemset/problem/1/A). Now it's finally getting repaved.

The square still has a rectangular shape of n × m meters. However, the picture is about to get more complicated now. Let a_{i,j} be the j-th square in the i-th row of the pavement.

You are given the picture of the squares:

  * if a_{i,j} =  "*", then the j-th square in the i-th row should be black; 
  * if a_{i,j} =  ".", then the j-th square in the i-th row should be white. 



The black squares are paved already. You have to pave the white squares. There are two options for pavement tiles:

  * 1 × 1 tiles — each tile costs x burles and covers exactly 1 square; 
  * 1 × 2 tiles — each tile costs y burles and covers exactly 2 adjacent squares of the same row. Note that you are not allowed to rotate these tiles or cut them into 1 × 1 tiles.



You should cover all the white squares, no two tiles should overlap and no black squares should be covered by tiles.

What is the smallest total price of the tiles needed to cover all the white squares?

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of testcases. Then the description of t testcases follow.

The first line of each testcase contains four integers n, m, x and y (1 ≤ n ≤ 100; 1 ≤ m ≤ 1000; 1 ≤ x, y ≤ 1000) — the size of the Theatre square, the price of the 1 × 1 tile and the price of the 1 × 2 tile.

Each of the next n lines contains m characters. The j-th character in the i-th line is a_{i,j}. If a_{i,j} =  "*", then the j-th square in the i-th row should be black, and if a_{i,j} =  ".", then the j-th square in the i-th row should be white.

It's guaranteed that the sum of n × m over all testcases doesn't exceed 10^5.

Output

For each testcase print a single integer — the smallest total price of the tiles needed to cover all the white squares in burles.

Example

Input


4
1 1 10 1
.
1 2 10 1
..
2 1 10 1
.
.
3 3 3 7
..*
*..
.*.


Output


10
1
20
18

Note

In the first testcase you are required to use a single 1 × 1 tile, even though 1 × 2 tile is cheaper. So the total price is 10 burles.

In the second testcase you can either use two 1 × 1 tiles and spend 20 burles or use a single 1 × 2 tile and spend 1 burle. The second option is cheaper, thus the answer is 1.

The third testcase shows that you can't rotate 1 × 2 tiles. You still have to use two 1 × 1 tiles for the total price of 20.

In the fourth testcase the cheapest way is to use 1 × 1 tiles everywhere. The total cost is 6 ⋅ 3 = 18.
import sys,bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
def Golf():n,*t=map(int,open(0).read().split())
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RA():return map(int,open(0).read().split())
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:
            a,b=inp;c=1
        else:
            a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in range(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def bit_combination(n,base=2):
    rt=[]
    for tb in range(base**n):s=[tb//(base**bt)%base for bt in range(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)
 
YN=['YES','NO'];Yn=['Yes','No']
mo=10**9+7
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**7)
read=sys.stdin.buffer.read
readline=sys.stdin.buffer.readline
input=lambda: sys.stdin.readline().rstrip()

class Tree:
    def __init__(self,inp_size=None,ls=None,init=True,index=0):
        self.LCA_init_stat=False
        self.ETtable=[]
        if init:
            if ls==None:
                self.stdin(inp_size,index=index)
            else:
                self.size=len(ls)+1
                self.edges,_=GI(self.size,self.size-1,ls,index=index)
        return
 
    def stdin(self,inp_size=None,index=1):
        if inp_size==None:
            self.size=int(input())
        else:
            self.size=inp_size
        self.edges,_=GI(self.size,self.size-1,index=index)
        return
    
    def listin(self,ls,index=0):
        self.size=len(ls)+1
        self.edges,_=GI(self.size,self.size-1,ls,index=index)
        return
 
    def __str__(self):
        return  str(self.edges)
 
    def dfs(self,x,func=lambda prv,nx,dist:prv+dist,root_v=0):
        q=deque()
        q.append(x)
        v=[-1]*self.size
        v[x]=root_v
        while q:
            c=q.pop()
            for nb,d in self.edges[c]:
                if v[nb]==-1:
                    q.append(nb)
                    v[nb]=func(v[c],nb,d)
        return v
 
    def EulerTour(self,x):
        q=deque()
        q.append(x)
        self.depth=[None]*self.size
        self.depth[x]=0
        self.ETtable=[]
        self.ETdepth=[]
        self.ETin=[-1]*self.size
        self.ETout=[-1]*self.size
        cnt=0
        while q:
            c=q.pop()
            if c<0:
                ce=~c
            else:
                ce=c
                for nb,d in self.edges[ce]:
                    if self.depth[nb]==None:
                        q.append(~ce)
                        q.append(nb)
                        self.depth[nb]=self.depth[ce]+1
            self.ETtable.append(ce)
            self.ETdepth.append(self.depth[ce])
            if self.ETin[ce]==-1:
                self.ETin[ce]=cnt
            else:
                self.ETout[ce]=cnt
            cnt+=1
        return
    
    def LCA_init(self,root):
        self.EulerTour(root)
        self.st=SparseTable(self.ETdepth,init_func=min,init_idl=inf)
        #self.st=SegTree(self.size*2-1,self.ETdepth,function=min,ide=inf)
        self.LCA_init_stat=True
        return
    
    def LCA(self,root,x,y):
        if self.LCA_init_stat==False:
            self.LCA_init(root)
        xin,xout=self.ETin[x],self.ETout[x]
        yin,yout=self.ETin[y],self.ETout[y]
        a=min(xin,yin)
        b=max(xout,yout,xin,yin)
        id_of_min_dep_in_et=self.st.query_id(a,b+1)
        return self.ETtable[id_of_min_dep_in_et]
 
class SparseTable: # O(N log N) for init, O(1) for query(l,r)
    def __init__(self,ls,init_func=min,init_idl=float('inf')):
        self.func=init_func
        self.idl=init_idl
        self.size=len(ls)
        self.N0=self.size.bit_length()
        self.table=[ls[:]]
        self.index=[list(range(self.size))]
        self.lg=[0]*(self.size+1)
        
        for i in range(2,self.size+1):
            self.lg[i]=self.lg[i>>1]+1
 
        for i in range(self.N0):
            tmp=[self.func(self.table[i][j],self.table[i][min(j+(1<<i),self.size-1)]) for j in range(self.size)]
            tmp_id=[self.index[i][j] if self.table[i][j]==self.func(self.table[i][j],self.table[i][min(j+(1<<i),self.size-1)]) else self.index[i][min(j+(1<<i),self.size-1)] for j in range(self.size)]
            self.table+=[tmp]
            self.index+=[tmp_id]
    
    # return func of [l,r)
    def query(self,l,r):
        if r>self.size:r=self.size
        #N=(r-l).bit_length()-1
        N=self.lg[r-l]
        return self.func(self.table[N][l],self.table[N][max(0,r-(1<<N))])
    
    # return index of which val[i] = func of v among [l,r)
    def query_id(self,l,r):
        if r>self.size:r=self.size
        #N=(r-l).bit_length()-1
        N=self.lg[r-l]
        a,b=self.index[N][l],self.index[N][max(0,r-(1<<N))]
        if self.table[0][a]==self.func(self.table[N][l],self.table[N][max(0,r-(1<<N))]):
            b=a
        return b
    
    def __str__(self):
        return str(self.table[0])
 
    def print(self):
        for i in self.table:
            print(*i)

show_flg=False
show_flg=True
ans=0

def solve():
    return

T=I()
for _ in range(T):
    n,m,x,y=LI()
    t=[]
    ans=0
    for i in range(n):
        s=[1 if i=='*' else 0 for i in input()]
        t+=[s]
    if y>2*x:
        y=x*2
        
    
    for i in range(n):
        for j in range(m):
            if t[i][j]==0:
                if j<m-1 and t[i][j+1]==0:
                    ans+=y
                    t[i][j]=1
                    t[i][j+1]=1
                else:
                    ans+=x
                    t[i][j]=1
    
    print(ans)
    

