#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int infi = INT32_MAX / 2;
const ll infl = INT64_MAX / 2;
const double pi = acos(-1);
vector<int> ans;
class Solve {
  string s;
  void set_value() { cin >> s; }
  int change;
  void judge() {
    change = 0;
    for (int i = (1); i < (s.size()); i++) {
      if (s[i] != s[i - 1]) ++change;
    }
  }
  void push() {
    if (change == 0)
      ans.push_back(1 - (s[0] - '0'));
    else if (change == 1)
      ans.push_back(1);
    else if (change == 2) {
      if (s[0] == '0')
        ans.push_back(2);
      else
        ans.push_back(1);
    } else
      ans.push_back(2);
  }

 public:
  Solve() : s(""), change(0){};
  ~Solve(){};
  void solve() {
    set_value();
    judge();
    push();
  }
};
int main() {
  int t;
  cin >> t;
  Solve problem;
  for (int i = (0); i < (t); i++) problem.solve();
  for (auto x : ans) cout << (x) << ("\n");
}

### Prompt

Create a solution in cpp for 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;
const int infi = INT32_MAX / 2;
const ll infl = INT64_MAX / 2;
const double pi = acos(-1);
vector<int> ans;
class Solve {
  string s;
  void set_value() { cin >> s; }
  int change;
  void judge() {
    change = 0;
    for (int i = (1); i < (s.size()); i++) {
      if (s[i] != s[i - 1]) ++change;
    }
  }
  void push() {
    if (change == 0)
      ans.push_back(1 - (s[0] - '0'));
    else if (change == 1)
      ans.push_back(1);
    else if (change == 2) {
      if (s[0] == '0')
        ans.push_back(2);
      else
        ans.push_back(1);
    } else
      ans.push_back(2);
  }

 public:
  Solve() : s(""), change(0){};
  ~Solve(){};
  void solve() {
    set_value();
    judge();
    push();
  }
};
int main() {
  int t;
  cin >> t;
  Solve problem;
  for (int i = (0); i < (t); i++) problem.solve();
  for (auto x : ans) cout << (x) << ("\n");
}

```