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 bisect
import math
import heapq
import itertools
import sys
from collections import deque
from atexit import register
from collections import Counter
from functools import reduce
if sys.version_info[0] < 3:
    from io import BytesIO as stream
else:
    from io import StringIO as stream
 
 
if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)
 
        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)
 
        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)
 
    input = raw_input
    range = xrange
 
    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip
 
 
def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.
 
    Args:
        sync (bool, optional): The new synchronization setting.
 
    """
    global input, flush
 
    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')
 
        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

def main():
    # sys.stdin  = open("input.txt")
    for _ in range(int(input())):
        n=int(input())
        s=input()
        cnt=1
        prev=s[0]
        v=[]
        for i in range(1,n):
            if s[i]==prev:
                cnt+=1
            else:
                v.append(cnt)
                cnt=1
                prev=s[i]
        v.append(cnt)
        n=len(v)
        arr=[0]*(n+5)
        last=-1
        for i in range(n-1,-1,-1):
            arr[i]=last
            if v[i]!=1:
                last=i
        a=last
        cnt=0
        i=0
        while a!=-1:
            v[a]-=1
            if v[a]==1 or i==a:
                a=arr[a]
            cnt+=1
            i+=1
        cnt+=int(math.ceil((n-i)/2))
        print(cnt)


                
                    

            
            


            

if __name__ == '__main__':
    sync_with_stdio(False)
    main()