#include <bits/stdc++.h>
const int INF = 1e9 + 5;
const int MOD = 1e9 + 7;
const double P = 3.141592;
const int hsh = 29;
using namespace std;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int ans = 0;
    int start = 0, togsgol = s.size() - 1;
    int barg = s.size();
    while (start < s.size() && togsgol >= 0 && barg > 0) {
      bool plus = false;
      for (start; start < s.size();) {
        if (s[start] == '1' && barg > 0) {
          start++;
          barg--;
        } else
          break;
      }
      for (togsgol; togsgol >= 0;) {
        if (s[togsgol] == '1' && barg > 0) {
          togsgol--;
          barg--;
        } else
          break;
      }
      while (barg > 0 && s[start] == '0') {
        start++;
        barg--;
        plus = true;
      }
      if (plus) ans++;
    }
    ans = min(ans, 2);
    cout << ans << '\n';
  }
}

### Prompt

Please create a solution in CPP 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>
const int INF = 1e9 + 5;
const int MOD = 1e9 + 7;
const double P = 3.141592;
const int hsh = 29;
using namespace std;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int ans = 0;
    int start = 0, togsgol = s.size() - 1;
    int barg = s.size();
    while (start < s.size() && togsgol >= 0 && barg > 0) {
      bool plus = false;
      for (start; start < s.size();) {
        if (s[start] == '1' && barg > 0) {
          start++;
          barg--;
        } else
          break;
      }
      for (togsgol; togsgol >= 0;) {
        if (s[togsgol] == '1' && barg > 0) {
          togsgol--;
          barg--;
        } else
          break;
      }
      while (barg > 0 && s[start] == '0') {
        start++;
        barg--;
        plus = true;
      }
      if (plus) ans++;
    }
    ans = min(ans, 2);
    cout << ans << '\n';
  }
}

```