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.
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 100;
inline int read() {
  int X = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    X = (X << 1) + (X << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return X;
  return ~(X - 1);
}
char s[N];
int main() {
  int _;
  cin >> _;
  while (_--) {
    cin >> s;
    int n = strlen(s);
    int ans = 0, f = 0;
    for (int i = 0; i <= n; i++) {
      if (s[i] == '0') f = 1;
    }
    if (f == 0) {
      cout << 0 << endl;
      continue;
    }
    int start = 0, sum = 0;
    for (int i = start; i < n; i++) {
      if (s[i] == '0') {
        int res = 0, yi = 0, er = 0;
        for (int j = i; j < n; j++) {
          if (s[j] == '1') yi++;
          if (s[j] == '0') res++;
          if (s[j] == '2') er++;
          start = j + 1;
          if (yi && er) break;
        }
        if (res >= 2 && yi == 1)
          sum++;
        else if (res >= 2 && er == 1)
          sum += 2;
        else if (res >= 2 && er == 0 && yi)
          sum += 2;
        else if (res >= 2 && yi == 0 && er)
          sum++;
        else
          sum++;
        i = start;
      }
    }
    cout << sum << endl;
  }
  return 0;
}
