#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("sse4.2")
void err(istringstream &ss) { cout << "\n"; }
template <typename T, typename... Args>
void err(istringstream &ss, const T &a, const Args &...args) {
  string token;
  getline(ss, token, ',');
  cout << token << " = " << a << "; ";
  err(ss, args...);
}
int log(int x) { return 32 - __builtin_clz(x) - 1; }
void read() {}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  cin >> t;
  read(args...);
}
struct Complex {
  double x, y;
  Complex(double x = 0, double y = 0) : x(x), y(y) {}
  Complex operator+(const Complex &a) { return Complex(x + a.x, y + a.y); }
  Complex operator-(const Complex &a) { return Complex(x - a.x, y - a.y); }
  Complex operator*(const Complex &a) {
    return Complex(x * a.x - y * a.y, x * a.y + y * a.x);
  }
};
bool check_palindrome(string s, int i, int n) {
  if (i >= n) return true;
  return (s[i] == s[n] && check_palindrome(s, i + 1, n - 1));
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
void solve() {
  string s;
  cin >> s;
  if (count(s.begin(), s.end(), '0') == 0) {
    cout << 0 << "\n";
  } else {
    if (int(s.rfind('0') - s.find('0') + 1) == count(s.begin(), s.end(), '0'))
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int t = 1;
  cin >> t;
  for (int c = 1; c <= t; c++) {
    solve();
  }
  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;
#pragma GCC target("sse4.2")
void err(istringstream &ss) { cout << "\n"; }
template <typename T, typename... Args>
void err(istringstream &ss, const T &a, const Args &...args) {
  string token;
  getline(ss, token, ',');
  cout << token << " = " << a << "; ";
  err(ss, args...);
}
int log(int x) { return 32 - __builtin_clz(x) - 1; }
void read() {}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  cin >> t;
  read(args...);
}
struct Complex {
  double x, y;
  Complex(double x = 0, double y = 0) : x(x), y(y) {}
  Complex operator+(const Complex &a) { return Complex(x + a.x, y + a.y); }
  Complex operator-(const Complex &a) { return Complex(x - a.x, y - a.y); }
  Complex operator*(const Complex &a) {
    return Complex(x * a.x - y * a.y, x * a.y + y * a.x);
  }
};
bool check_palindrome(string s, int i, int n) {
  if (i >= n) return true;
  return (s[i] == s[n] && check_palindrome(s, i + 1, n - 1));
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
void solve() {
  string s;
  cin >> s;
  if (count(s.begin(), s.end(), '0') == 0) {
    cout << 0 << "\n";
  } else {
    if (int(s.rfind('0') - s.find('0') + 1) == count(s.begin(), s.end(), '0'))
      cout << 1 << "\n";
    else
      cout << 2 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int t = 1;
  cin >> t;
  for (int c = 1; c <= t; c++) {
    solve();
  }
  return 0;
}

```