#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ","
       << " ";
  err(++it, args...);
  cout << "\n";
}
bool checkbit(long long mask, long long bit) { return mask & (1ll << bit); }
int setbit(long long mask, long long bit) { return mask | (1ll << bit); }
int clearbit(long long mask, long long bit) { return mask & ~(1ll << bit); }
int togglebit(long long mask, long long bit) { return mask ^ (1ll << bit); }
int bitno(long long mask) { return (long long)__builtin_popcount(mask); }
const int fx[] = {+1, -1, +0, +0};
const int fy[] = {+0, +0, +1, -1};
const long long maX = 1e6 + 10;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long test;
  cin >> test;
  while (test--) {
    string str;
    cin >> str;
    long long n = str.size();
    long long cc = 0;
    for (long long i = 0; i + 1 < n; i++) {
      if (str[i] == '0' && str[i] != str[i + 1]) cc++;
    }
    if (str[n - 1] == '0') cc++;
    cout << min(cc, 2ll) << "\n";
  }
  return 0;
}

### 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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ","
       << " ";
  err(++it, args...);
  cout << "\n";
}
bool checkbit(long long mask, long long bit) { return mask & (1ll << bit); }
int setbit(long long mask, long long bit) { return mask | (1ll << bit); }
int clearbit(long long mask, long long bit) { return mask & ~(1ll << bit); }
int togglebit(long long mask, long long bit) { return mask ^ (1ll << bit); }
int bitno(long long mask) { return (long long)__builtin_popcount(mask); }
const int fx[] = {+1, -1, +0, +0};
const int fy[] = {+0, +0, +1, -1};
const long long maX = 1e6 + 10;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long test;
  cin >> test;
  while (test--) {
    string str;
    cin >> str;
    long long n = str.size();
    long long cc = 0;
    for (long long i = 0; i + 1 < n; i++) {
      if (str[i] == '0' && str[i] != str[i + 1]) cc++;
    }
    if (str[n - 1] == '0') cc++;
    cout << min(cc, 2ll) << "\n";
  }
  return 0;
}

```