#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void inout() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  inout();
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int c0 = 0;
    long long int c1 = 0;
    long long int n = s.size();
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '0')
        c0++;
      else
        c1++;
    }
    if (c0 == n) {
      cout << 1 << "\n";
      continue;
    }
    if (c1 == n) {
      cout << 0 << "\n";
      continue;
    }
    long long int c = 0;
    char ch = s[0];
    for (long long int i = 1; i < n; i++) {
      if (s[i] == ch) continue;
      if (ch == '0') {
        ch = s[i];
        c++;
      } else {
        ch = s[i];
      }
    }
    if (ch == '0') c++;
    if (c > 2)
      cout << 2 << "\n";
    else
      cout << c << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void inout() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  inout();
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int c0 = 0;
    long long int c1 = 0;
    long long int n = s.size();
    for (long long int i = 0; i < n; i++) {
      if (s[i] == '0')
        c0++;
      else
        c1++;
    }
    if (c0 == n) {
      cout << 1 << "\n";
      continue;
    }
    if (c1 == n) {
      cout << 0 << "\n";
      continue;
    }
    long long int c = 0;
    char ch = s[0];
    for (long long int i = 1; i < n; i++) {
      if (s[i] == ch) continue;
      if (ch == '0') {
        ch = s[i];
        c++;
      } else {
        ch = s[i];
      }
    }
    if (ch == '0') c++;
    if (c > 2)
      cout << 2 << "\n";
    else
      cout << c << "\n";
  }
  return 0;
}

```