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;
  using namespace std::chrono;
  template <typename T, typename U>
  long long modMul(T a, U b) {
    return ((a % 1000000007) * 1LL * (b % 1000000007)) % 1000000007;
  }
  template <typename T, typename U>
  long long binpower(T n, U k) {
    T res = 1;
    while (k) {
      if (k & 1) {
        res = modMul(res, n);
      }
      n = modMul(n, n);
      k >>= 1;
    }
    return res % 1000000007;
  }
  template <typename T>
  T moduloInverse(T a) {
    return binpower(a, 1000000007 - 2);
  }
  template <typename T, typename U>
  long long modAdd(T a, U b) {
    return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
  }
  template <typename T, typename U>
  long long modSub(T a, U b) {
    return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
  }
  template <typename T, typename U>
  long long modDiv(T a, U b) {
    return modMul(a, moduloInverse(b));
  }
  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 / gcd(a, b)) * b; }
  void _print(long long x) { cerr << x; }
  void _print(float x) { cerr << x; }
  void _print(double x) { cerr << x; }
  void _print(string x) { cerr << x; }
  void _print(unsigned long long x) { cerr << x; }
  void _print(char x) { cerr << x; }
  void _print(bool x) { cerr << x; }
  template <typename T>
  vector<T> inp(long long n) {
    vector<T> ain(n);
    for (long long i = 0; i < n; i++) cin >> ain[i];
    return ain;
  }
  template <typename T>
  void print(vector<T> arr) {
    for (long long i = 0; i < arr.size(); i++) cout << arr[i] << " ";
    cout << "\n";
    return;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> arr = inp<long long>(n);
    vector<long long> brr;
    long long h = 1;
    long long prev = -1;
    long long ct = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i]) {
        if (i > 0) {
          if (arr[i - 1]) {
            h += 5;
          } else {
            h += 1;
          }
        } else {
          h++;
        }
      } else {
        if (i > 0) {
          if (arr[i - 1] == 0) {
            h = -1;
            break;
          }
        }
      }
    }
    cout << h << "\n";
    return;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  