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;
template <class T>
bool chkmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
long long read1() {
  long long num = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num *= 10;
    num += ch - '0';
    ch = getchar();
  }
  return num * f;
}
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
int idx(char x) { return x - 'a'; }
long long n, k;
int a[300010];
int b[10000];
int ans[100][100];
map<int, int> id;
int main() {
  int t;
  t = read();
  while (t--) {
    string s;
    getline(cin, s);
    long long ans = 0;
    bool flag = false;
    for (int i = 0; i <= s.size(); i++) {
      if (s[i] == '1') {
        ans += flag;
        flag = false;
      } else if (s[i] == '0')
        flag = true;
    }
    if (flag) ans++;
    if (ans > 2) ans = 2;
    cout << ans << endl;
  }
  return 0;
}
