#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vl = vector<ll>;
using pll = pair<ll, ll>;
using qll = queue<ll>;
using vp = vector<pll>;
using qp = queue<pll>;
using pqll = priority_queue<ll>;
using pqp = priority_queue<pll>;
inline void init_setting() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
int t;
string s;
int main() {
  init_setting();
  cin >> t;
  while (t--) {
    cin >> s;
    ll n = s.size();
    ll check = s[0] == '0';
    for (int i = 1; i <= n - 1; i++)
      if (s[i] == '0' && s[i - 1] != s[i]) check++;
    if (check < 2)
      cout << check;
    else
      cout << 2;
    cout << '\n';
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

A binary string is a string that consists of characters 0 and 1.

Let \operatorname{MEX} of a binary string be the smallest digit among 0, 1, or 2 that does not occur in the string. For example, \operatorname{MEX} of 001011 is 2, because 0 and 1 occur in the string at least once, \operatorname{MEX} of 1111 is 0, because 0 and 2 do not occur in the string and 0 < 2.

A binary string s is given. You should cut it into any number of substrings such that each character is in exactly one substring. It is possible to cut the string into a single substring — the whole string.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

What is the minimal sum of \operatorname{MEX} of all substrings pieces can be?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

Each test case contains a single binary string s (1 ≤ |s| ≤ 10^5).

It's guaranteed that the sum of lengths of s over all test cases does not exceed 10^5.

Output

For each test case print a single integer — the minimal sum of \operatorname{MEX} of all substrings that it is possible to get by cutting s optimally.

Example

Input


6
01
1111
01100
101
0000
01010


Output


1
0
2
1
1
2

Note

In the first test case the minimal sum is \operatorname{MEX}(0) + \operatorname{MEX}(1) = 1 + 0 = 1.

In the second test case the minimal sum is \operatorname{MEX}(1111) = 0.

In the third test case the minimal sum is \operatorname{MEX}(01100) = 2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vl = vector<ll>;
using pll = pair<ll, ll>;
using qll = queue<ll>;
using vp = vector<pll>;
using qp = queue<pll>;
using pqll = priority_queue<ll>;
using pqp = priority_queue<pll>;
inline void init_setting() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
int t;
string s;
int main() {
  init_setting();
  cin >> t;
  while (t--) {
    cin >> s;
    ll n = s.size();
    ll check = s[0] == '0';
    for (int i = 1; i <= n - 1; i++)
      if (s[i] == '0' && s[i - 1] != s[i]) check++;
    if (check < 2)
      cout << check;
    else
      cout << 2;
    cout << '\n';
  }
}

```