#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long remi = 1000000007;
const long long remi2 = 998244353;
const long long inf = 1e18 + 1e17 + 1e16;
void XORing() {
  string s;
  cin >> s;
  long long c0 = 0, c1 = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0') {
      c0++;
    } else if (s[i] == '1') {
      c1++;
    }
  }
  if (c0 == 0) {
    cout << 0 << "\n";
    return;
  } else if (c1 == 0) {
    cout << 1 << "\n";
    return;
  }
  long long as1, as2 = 0, as3 = 0;
  if (c1 > 0 and c0 > 0) {
    as1 = 2;
  }
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0') {
      as2 += 1;
    }
  }
  long long i, j;
  for (i = 0; i < s.size(); i = j) {
    char c = s[i];
    for (j = i + 1; j < s.size(); j++) {
      if (s[j] == c) {
        continue;
      } else {
        break;
      }
    }
    if (c == '0') {
      as3 += 1;
    }
  }
  cout << min(as1, min(as2, as3)) << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  for (int i = 1; i < T + 1; i++) {
    XORing();
  }
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long remi = 1000000007;
const long long remi2 = 998244353;
const long long inf = 1e18 + 1e17 + 1e16;
void XORing() {
  string s;
  cin >> s;
  long long c0 = 0, c1 = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0') {
      c0++;
    } else if (s[i] == '1') {
      c1++;
    }
  }
  if (c0 == 0) {
    cout << 0 << "\n";
    return;
  } else if (c1 == 0) {
    cout << 1 << "\n";
    return;
  }
  long long as1, as2 = 0, as3 = 0;
  if (c1 > 0 and c0 > 0) {
    as1 = 2;
  }
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == '0') {
      as2 += 1;
    }
  }
  long long i, j;
  for (i = 0; i < s.size(); i = j) {
    char c = s[i];
    for (j = i + 1; j < s.size(); j++) {
      if (s[j] == c) {
        continue;
      } else {
        break;
      }
    }
    if (c == '0') {
      as3 += 1;
    }
  }
  cout << min(as1, min(as2, as3)) << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  for (int i = 1; i < T + 1; i++) {
    XORing();
  }
  return 0;
}

```