description: You are given a digital clock with n digits. Each digit shows an integer from 0 to 9, so the whole clock shows an integer from 0 to 10^n-1. The clock will show leading zeroes if the number is smaller than 10^{n-1}.
  
  You want the clock to show 0 with as few operations as possible. In an operation, you can do one of the following: 
  
    * decrease the number on the clock by 1, or 
    * swap two digits (you can choose which digits to swap, and they don't have to be adjacent). 
  
  
  
  Your task is to determine the minimum number of operations needed to make the clock show 0.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3).
  
  The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — number of digits on the clock.
  
  The second line of each test case contains a string of n digits s_1, s_2, …, s_n (0 ≤ s_1, s_2, …, s_n ≤ 9) — the number on the clock.
  
  Note: If the number is smaller than 10^{n-1} the clock will show leading zeroes.
  
  Output
  
  For each test case, print one integer: the minimum number of operations needed to make the clock show 0.
  
  Example
  
  Input
  
  
  7
  3
  007
  4
  1000
  5
  00000
  3
  103
  4
  2020
  9
  123456789
  30
  001678294039710047203946100020
  
  
  Output
  
  
  7
  2
  0
  5
  6
  53
  115
  
  Note
  
  In the first example, it's optimal to just decrease the number 7 times.
  
  In the second example, we can first swap the first and last position and then decrease the number by 1.
  
  In the third example, the clock already shows 0, so we don't have to perform any operations.
solution: for _ in range(int(input())):
      n=int(input())
      s=int(input())
      s=str(s)
      ans=0
      for i in range(len(s)-1,-1,-1):
          if int(s[i])!=0 and i==len(s)-1:
              ans+=int(s[i])
          elif int(s[i])!=0:
              ans+=(1+int(s[i]))
      print(ans)