#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int findGCD(int arr[], int n) {
  int result = arr[0];
  for (int i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if ((y % 2) == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    int ct = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == '0') {
        ct++;
      }
    }
    int ct0 = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == '1' && s[i + 1] == '0') {
        ct0++;
      }
    }
    if (s[0] == '0') {
      ct0++;
    }
    if (ct == 0) {
      cout << 0;
    } else if (ct == n) {
      cout << 1;
    } else if (ct0 > 2) {
      cout << 2;
    } else {
      cout << ct0;
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int findGCD(int arr[], int n) {
  int result = arr[0];
  for (int i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if ((y % 2) == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    int ct = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == '0') {
        ct++;
      }
    }
    int ct0 = 0;
    for (int i = 0; i <= n - 1; i++) {
      if (s[i] == '1' && s[i + 1] == '0') {
        ct0++;
      }
    }
    if (s[0] == '0') {
      ct0++;
    }
    if (ct == 0) {
      cout << 0;
    } else if (ct == n) {
      cout << 1;
    } else if (ct0 > 2) {
      cout << 2;
    } else {
      cout << ct0;
    }
    cout << "\n";
  }
  return 0;
}

```