#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using ull = unsigned long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using vs = vector<string>;
using vvs = vector<vs>;
using vc = vector<char>;
using vvc = vector<vc>;
using ii = pair<int, int>;
using vii = vector<ii>;
using vvii = vector<vii>;
using dd = pair<double, double>;
using vdd = vector<dd>;
using vvdd = vector<vdd>;
using Mii = map<int, int>;
using vMii = vector<Mii>;
using Si = set<int>;
using vSi = vector<Si>;
using vvSi = vector<vSi>;
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    bool one = true;
    for (char c : s)
      if (c == '0') one = false;
    if (one)
      cout << 0 << '\n';
    else {
      int first = -1;
      for (int i = 0; i < n; ++i) {
        if (s[i] == '0') {
          first = i;
          break;
        }
      }
      int last = -1;
      for (int i = n - 1; i >= 0; --i) {
        if (s[i] == '0') {
          last = i;
          break;
        }
      }
      bool two = false;
      for (int i = first; i <= last; ++i) {
        if (s[i] == '1') two = true;
      }
      cout << (two ? 2 : 1) << '\n';
    }
  }
}

### Prompt

Please formulate 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>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using ull = unsigned long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vvd = vector<vd>;
using vvvd = vector<vvd>;
using vs = vector<string>;
using vvs = vector<vs>;
using vc = vector<char>;
using vvc = vector<vc>;
using ii = pair<int, int>;
using vii = vector<ii>;
using vvii = vector<vii>;
using dd = pair<double, double>;
using vdd = vector<dd>;
using vvdd = vector<vdd>;
using Mii = map<int, int>;
using vMii = vector<Mii>;
using Si = set<int>;
using vSi = vector<Si>;
using vvSi = vector<vSi>;
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    bool one = true;
    for (char c : s)
      if (c == '0') one = false;
    if (one)
      cout << 0 << '\n';
    else {
      int first = -1;
      for (int i = 0; i < n; ++i) {
        if (s[i] == '0') {
          first = i;
          break;
        }
      }
      int last = -1;
      for (int i = n - 1; i >= 0; --i) {
        if (s[i] == '0') {
          last = i;
          break;
        }
      }
      bool two = false;
      for (int i = first; i <= last; ++i) {
        if (s[i] == '1') two = true;
      }
      cout << (two ? 2 : 1) << '\n';
    }
  }
}

```