#include <bits/stdc++.h>
long long mod = 1000000007;
using namespace std;
void solve() {
  string s;
  cin >> s;
  int left = 0;
  int right = s.size() - 1;
  int p1, p2;
  while (left < s.size() - 1 && s[left] == s[left + 1]) {
    left++;
  }
  if (left == s.size() - 1 && s[left] == s[right]) {
    int c1 = (s[left] - 48) ^ 1;
    cout << c1 << endl;
    return;
  }
  while (right > 1 && s[right] == s[right - 1]) {
    right--;
  }
  if (right == 0 && s[left] == s[right]) {
    int c2 = (s[right] - 48) ^ 1;
    cout << c2 << endl;
    return;
  }
  p1 = left + 1;
  p2 = right - 1;
  bool zero = false, one = false;
  while (p1 <= p2) {
    if (s[p1] == '0' || s[p2] == '0') {
      zero = true;
    }
    if (s[p1] == '1' || s[p2] == '1') {
      one = true;
    }
    p1++;
    p2--;
  }
  int three = 0;
  if (zero && one) {
    three = 2;
  } else if (zero) {
    three = 1;
  } else if (one) {
    three = 0;
  } else {
    int c1 = (s[left] - 48) ^ 1;
    int c2 = (s[right] - 48) ^ 1;
    cout << min(2, c1 + c2 + three) << endl;
    return;
  }
  int c1 = (s[left] - 48) ^ 1;
  int c2 = (s[right] - 48) ^ 1;
  cout << min(2, c1 + c2 + three) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
long long mod = 1000000007;
using namespace std;
void solve() {
  string s;
  cin >> s;
  int left = 0;
  int right = s.size() - 1;
  int p1, p2;
  while (left < s.size() - 1 && s[left] == s[left + 1]) {
    left++;
  }
  if (left == s.size() - 1 && s[left] == s[right]) {
    int c1 = (s[left] - 48) ^ 1;
    cout << c1 << endl;
    return;
  }
  while (right > 1 && s[right] == s[right - 1]) {
    right--;
  }
  if (right == 0 && s[left] == s[right]) {
    int c2 = (s[right] - 48) ^ 1;
    cout << c2 << endl;
    return;
  }
  p1 = left + 1;
  p2 = right - 1;
  bool zero = false, one = false;
  while (p1 <= p2) {
    if (s[p1] == '0' || s[p2] == '0') {
      zero = true;
    }
    if (s[p1] == '1' || s[p2] == '1') {
      one = true;
    }
    p1++;
    p2--;
  }
  int three = 0;
  if (zero && one) {
    three = 2;
  } else if (zero) {
    three = 1;
  } else if (one) {
    three = 0;
  } else {
    int c1 = (s[left] - 48) ^ 1;
    int c2 = (s[right] - 48) ^ 1;
    cout << min(2, c1 + c2 + three) << endl;
    return;
  }
  int c1 = (s[left] - 48) ^ 1;
  int c2 = (s[right] - 48) ^ 1;
  cout << min(2, c1 + c2 + three) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}

```