You have a string s consisting of n characters. Each character is either 0 or 1.

You can perform operations on the string. Each operation consists of two steps:

  1. select an integer i from 1 to the length of the string s, then delete the character s_i (the string length gets reduced by 1, the indices of characters to the right of the deleted one also get reduced by 1); 
  2. if the string s is not empty, delete the maximum length prefix consisting of the same characters (the indices of the remaining characters and the string length get reduced by the length of the deleted prefix). 



Note that both steps are mandatory in each operation, and their order cannot be changed.

For example, if you have a string s = 111010, the first operation can be one of the following:

  1. select i = 1: we'll get 111010 → 11010 → 010; 
  2. select i = 2: we'll get 111010 → 11010 → 010; 
  3. select i = 3: we'll get 111010 → 11010 → 010; 
  4. select i = 4: we'll get 111010 → 11110 → 0; 
  5. select i = 5: we'll get 111010 → 11100 → 00; 
  6. select i = 6: we'll get 111010 → 11101 → 01. 



You finish performing operations when the string s becomes empty. What is the maximum number of operations you can perform?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the string s.

The second line contains string s of n characters. Each character is either 0 or 1.

It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print a single integer — the maximum number of operations you can perform.

Example

Input


5
6
111010
1
0
1
1
2
11
6
101010


Output


3
1
1
1
3

Note

In the first test case, you can, for example, select i = 2 and get string 010 after the first operation. After that, you can select i = 3 and get string 1. Finally, you can only select i = 1 and get empty string.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
import sys
sys.setrecursionlimit(10**4)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
    for i in range(3,int(math.sqrt(n))+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            a.append(i) 
            n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        a.append(n)
    return a
def fun(p,q):
    a=[]
    a.append((1,-1,q-p))
    a.append((1,0,-p))
    a.append((0,1,-p))
    return a
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=sin()
        x=a[0]
        p=0
        ans=[]
        for i in a:
            if i==x:
                p+=1
            else:
                ans.append(p)
                x=i
                p=1
        ans.append(p)
        n=len(ans)
        i=j=h=0
        while(i<n):
            j=max(i,j)
            while(j<n and ans[j]==1):
                j+=1
            if j<n:
                ans[j]-=1
            else:
                i+=1
            i+=1
            h+=1
        print(h)


        
    
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is yferent which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()