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;
  #pragma GCC optimize("Ofast")
  const long long int maxn = 2e5 + 16, md = 1e9 + 7, inf = 2e16;
  long long int gcd(long long int a, long long int b) {
    if (a < b) swap(a, b);
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  inline long long int tav(long long int n, long long int k) {
    long long int res = 1;
    while (k > 0) {
      if (k & 1) {
        res *= n;
        res %= md;
      }
      n *= n;
      n %= md;
      k >>= 1;
    }
    return res;
  }
  long long int a[maxn];
  void solve() {
    long long int n, cur = 1;
    bool dead = false;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (i) {
        if (!a[i] && !a[i - 1]) {
          dead = true;
        }
      }
      cur += a[i];
      cur += 4 * (a[i] == 1 && a[i - 1] == 1);
    }
    if (dead) {
      cout << "-1\n";
      return;
    }
    cout << cur << '\n';
    return;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int T;
    cin >> T;
    while (T--) solve();
    return 0;
  }
  