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.
"""
    Author: Sagar Pandey

"""
# ---------------------------------------------------Import Libraries---------------------------------------------------
import sys
import os
from math import sqrt, log, log2, log10, gcd, floor, pow, sin, cos, tan, pi, inf, factorial
from copy import copy, deepcopy
from sys import exit, stdin, stdout
from collections import Counter, defaultdict, deque
from itertools import permutations
import heapq
from bisect import bisect_left as bl
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is r

# ---------------------------------------------------Global Variables---------------------------------------------------
# sys.setrecursionlimit(100000000)
mod = 1000000007
# ---------------------------------------------------Helper Functions---------------------------------------------------
iinp = lambda: int(sys.stdin.readline())
inp = lambda: sys.stdin.readline().strip()
strl = lambda: list(inp().strip().split(" "))
intl = lambda: list(map(int, inp().split(" ")))
mint = lambda: map(int, inp().split())
flol = lambda: list(map(float, inp().split(" ")))
flush = lambda: stdout.flush()


def permute(nums):
    def fun(arr, nums, cur, v):
        if len(cur) == len(nums):
            arr.append(cur.copy())
        i = 0
        while i < len(nums):
            if v[i]:
                i += 1
                continue
            else:
                cur.append(nums[i])
                v[i] = 1
                fun(arr, nums, cur, v)
                cur.pop()
                v[i] = 0
                i += 1
            # while i<len(nums) and nums[i]==nums[i-1]:i+=1    # Uncomment for unique permutations
        return arr

    res = []
    nums.sort()
    v = [0] * len(nums)
    return fun(res, nums, [], v)


def subsets(res, index, arr, cur):
    res.append(cur.copy())
    for i in range(index, len(arr)):
        cur.append(arr[i])
        subsets(res, i + 1, arr, cur)
        cur.pop()
    return res


def sieve(N):
    root = int(sqrt(N))
    primes = [1] * (N + 1)
    primes[0], primes[1] = 0, 0
    for i in range(2, root + 1):
        if primes[i]:
            for j in range(i * i, N + 1, i):
                primes[j] = 0
    return primes


def bs(arr, l, r, x):
    if x < arr[0] or x > arr[len(arr) - 1]:
        return -1
    while l <= r:
        mid = l + (r - l) // 2
        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            l = mid + 1
        else:
            r = mid - 1
    return -1


def isPrime(n):
    if n <= 1: return False
    if n <= 3: return True
    if n % 2 == 0 or n % 3 == 0: return False
    p = int(sqrt(n))
    for i in range(5, p + 1, 6):
        if n % i == 0 or n % (i + 2) == 0:
            return False
    return True


# -------------------------------------------------------Functions------------------------------------------------------

def solve():
    n = iinp()
    s = inp()
    seg = []
    count = 1
    for i in range(1, n):
        if s[i] == s[i - 1]:
            count += 1
        else:
            seg.append(count)
            count = 1
    seg.append(count)
    # print(seg)
    
    ans=0
    pt=0
    i=0
    n=len(seg)
    while i<n:
        while pt<n:
            if seg[pt]>1:
                seg[pt]-=1
                break
            pt+=1
        if pt==n:
            i+=1
        while pt<=i:
            pt+=1
        ans+=1
        i+=1
    print(ans)

# -------------------------------------------------------Main Code------------------------------------------------------
for _ in range(iinp()):
    solve()
