Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.

Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).

Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.

Output

For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].

Example

Input


5
4
37
998244353
1000000007
12345678901337426966631415


Output


1
2
9
10
26

Note

Explanations for the two first test cases of the example:

  1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1; 
  2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). 
#include<bits/stdc++.h>
#include<string>
using namespace std;
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
    fastio;
    string s;
    int t,n;
    cin>>t;
    while(t--)
    {
        cin>>s;
        n=s.size();
        cout<<n<<"\n";
    }
    return 0;
}
