description: Petya has got an interesting flower. Petya is a busy person, so he sometimes forgets to water it. You are given n days from Petya's live and you have to determine what happened with his flower in the end.
  
  The flower grows as follows: 
  
    * If the flower isn't watered for two days in a row, it dies. 
    * If the flower is watered in the i-th day, it grows by 1 centimeter. 
    * If the flower is watered in the i-th and in the (i-1)-th day (i > 1), then it grows by 5 centimeters instead of 1. 
    * If the flower is not watered in the i-th day, it does not grow. 
  
  
  
  At the beginning of the 1-st day the flower is 1 centimeter tall. What is its height after n days?
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The first line of each test case contains the only integer n (1 ≤ n ≤ 100).
  
  The second line of each test case contains n integers a_1, a_2, ..., a_n (a_i = 0 or a_i = 1). If a_i = 1, the flower is watered in the i-th day, otherwise it is not watered.
  
  Output
  
  For each test case print a single integer k — the flower's height after n days, or -1, if the flower dies.
  
  Example
  
  Input
  
  
  4
  3
  1 0 1
  3
  0 1 1
  4
  1 0 0 1
  1
  0
  
  
  Output
  
  
  3
  7
  -1
  1
solution: #include <bits/stdc++.h>
  using namespace std;
  const int M = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long inv(long long i, long long m) {
    if (i == 1) return 1;
    return (m - ((m / i) * inv(m % i, m)) % m) % m;
  }
  long long mod(long long x, long long m) { return ((x % m + m) % m); }
  long long mod_add(long long a, long long b, long long m) {
    return mod((mod(a, m) + mod(b, m)), m);
  }
  long long mod_sub(long long a, long long b, long long m) {
    return mod((mod(a, m) - mod(b, m)), m);
  }
  long long mod_mul(long long a, long long b, long long m) {
    return mod((mod(a, m) * mod(b, m)), m);
  }
  long long expo(long long a, long long b, long long m) {
    long long res = 1;
    while (b) {
      if (b & 1) {
        res = mod_mul(res, a, m);
      }
      a = mod_mul(a, a, m);
      b = b >> 1;
    }
    return res;
  }
  void swap(int &a, int &b) {
    a ^= b;
    b ^= a;
    a ^= b;
  }
  long long pwr(long long a, long long b, long long mod) {
    a %= mod;
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return res;
  }
  bool pow_of_2(long long n) {
    if (n != 0 && (n & n - 1) == 0) return true;
    return false;
  }
  void solve() {
    int n;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long sum = 1;
    for (int i = 0; i < n; i++) {
      if (arr[i] == 1) {
        if (i + 1 < n && arr[i + 1] == 1)
          sum += 5;
        else
          sum++;
      } else if (arr[i] == 0) {
        if (i + 1 < n && arr[i + 1] == 0) {
          cout << -1 << endl;
          return;
        }
      }
    }
    cout << sum << endl;
  }
  int main() {
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  