#include <bits/stdc++.h>
using namespace std;
int expo(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = (res * x % 1000000007) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nextPrime(long long N) {
  if (N <= 1) return 2;
  long long prime = N;
  bool found = false;
  while (!found) {
    prime++;
    if (isPrime(prime)) found = true;
  }
  return prime;
}
long long cl(long long n, long long d) { return (n + d - 1) / d; }
long long binToDec(string s) { return bitset<64>(s).to_ullong(); }
string decToBin(long long a) { return bitset<64>(a).to_string(); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
void print(pair<long long, long long> a) {
  cout << a.first << ' ' << a.second;
  cout << endl;
  ;
}
void print(long long a) {
  cout << a;
  cout << endl;
}
void print(std::vector<long long> a) {
  for (auto i : a) cout << i << ' ';
  cout << endl;
  ;
}
void print(string s) {
  cout << s;
  cout << endl;
  ;
}
void printr(std::vector<long long> a, long long start, long long end) {
  for (long long i = start; i < end; i++) cout << a[i] << ' ';
  cout << endl;
  ;
}
void print(long long a, long long b) { cout << a << ' ' << b; }
void amax(long long &a, long long b) { a = max(a, b); }
void amin(long long &a, long long b) { a = min(a, b); }
long long madd(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long mmul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long msub(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long fpow(long long x, long long y, long long p = 1000000007) {
  x %= p;
  long long sum = 1;
  while (y) {
    if (y & 1) sum = sum * x;
    sum %= p;
    y = y >> 1;
    x = x * x;
    x %= p;
  }
  return sum;
}
bool isPerSquare(long double a) {
  if (a < 0) return false;
  long long sr = sqrt(a);
  return (sr * sr == a);
}
bool isPalindrome(string str) {
  long long low = 0, high = long long(str.size()) - 1;
  while (low < high) {
    if (str[low] != str[high]) return false;
    low++, high--;
  }
  return true;
}
void solve() {
  long long n;
  cin >> n;
  ;
  string s;
  cin >> s;
  ;
  for (auto i : s) {
    if (!isPrime(i - '0')) {
      print(1);
      cout << i << '\n';
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      string num(1, s[i]);
      num += string(1, s[j]);
      long long nu = stoll(num);
      if (!isPrime(nu)) {
        print(2);
        print(nu);
        return;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Generate a Cpp solution to the following problem:

During the hypnosis session, Nicholas suddenly remembered a positive integer n, which doesn't contain zeros in decimal notation. 

Soon, when he returned home, he got curious: what is the maximum number of digits that can be removed from the number so that the number becomes not prime, that is, either composite or equal to one?

For some numbers doing so is impossible: for example, for number 53 it's impossible to delete some of its digits to obtain a not prime integer. However, for all n in the test cases of this problem, it's guaranteed that it's possible to delete some of their digits to obtain a not prime number.

Note that you cannot remove all the digits from the number.

A prime number is a number that has no divisors except one and itself. A composite is a number that has more than two divisors. 1 is neither a prime nor a composite number.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 10^3), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer k (1 ≤ k ≤ 50) — the number of digits in the number.

The second line of each test case contains a positive integer n, which doesn't contain zeros in decimal notation (10^{k-1} ≤ n < 10^{k}). It is guaranteed that it is always possible to remove less than k digits to make the number not prime.

It is guaranteed that the sum of k over all test cases does not exceed 10^4.

Output

For every test case, print two numbers in two lines. In the first line print the number of digits, that you have left in the number. In the second line print the digits left after all delitions. 

If there are multiple solutions, print any.

Example

Input


7
3
237
5
44444
3
221
2
35
3
773
1
4
30
626221626221626221626221626221


Output


2
27
1
4
1
1
2
35
2
77
1
4
1
6

Note

In the first test case, you can't delete 2 digits from the number 237, as all the numbers 2, 3, and 7 are prime. However, you can delete 1 digit, obtaining a number 27 = 3^3.

In the second test case, you can delete all digits except one, as 4 = 2^2 is a composite number.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int expo(int x, int y) {
  int res = 1;
  while (y) {
    if (y % 2) res = (res * x % 1000000007) % 1000000007;
    x = (x * x) % 1000000007;
    y /= 2;
  }
  return res;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nextPrime(long long N) {
  if (N <= 1) return 2;
  long long prime = N;
  bool found = false;
  while (!found) {
    prime++;
    if (isPrime(prime)) found = true;
  }
  return prime;
}
long long cl(long long n, long long d) { return (n + d - 1) / d; }
long long binToDec(string s) { return bitset<64>(s).to_ullong(); }
string decToBin(long long a) { return bitset<64>(a).to_string(); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
void print(pair<long long, long long> a) {
  cout << a.first << ' ' << a.second;
  cout << endl;
  ;
}
void print(long long a) {
  cout << a;
  cout << endl;
}
void print(std::vector<long long> a) {
  for (auto i : a) cout << i << ' ';
  cout << endl;
  ;
}
void print(string s) {
  cout << s;
  cout << endl;
  ;
}
void printr(std::vector<long long> a, long long start, long long end) {
  for (long long i = start; i < end; i++) cout << a[i] << ' ';
  cout << endl;
  ;
}
void print(long long a, long long b) { cout << a << ' ' << b; }
void amax(long long &a, long long b) { a = max(a, b); }
void amin(long long &a, long long b) { a = min(a, b); }
long long madd(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long mmul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long msub(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long fpow(long long x, long long y, long long p = 1000000007) {
  x %= p;
  long long sum = 1;
  while (y) {
    if (y & 1) sum = sum * x;
    sum %= p;
    y = y >> 1;
    x = x * x;
    x %= p;
  }
  return sum;
}
bool isPerSquare(long double a) {
  if (a < 0) return false;
  long long sr = sqrt(a);
  return (sr * sr == a);
}
bool isPalindrome(string str) {
  long long low = 0, high = long long(str.size()) - 1;
  while (low < high) {
    if (str[low] != str[high]) return false;
    low++, high--;
  }
  return true;
}
void solve() {
  long long n;
  cin >> n;
  ;
  string s;
  cin >> s;
  ;
  for (auto i : s) {
    if (!isPrime(i - '0')) {
      print(1);
      cout << i << '\n';
      return;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      string num(1, s[i]);
      num += string(1, s[j]);
      long long nu = stoll(num);
      if (!isPrime(nu)) {
        print(2);
        print(nu);
        return;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) solve();
}

```