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;
  std::vector<std::string> svec;
  vector<long long int> v;
  void sieve(long long int n) {
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
    for (long long int p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long int i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    for (long long int p = 2; p <= n; p++)
      if (prime[p]) v.push_back(p);
  }
  long long int gcd(long long int a, int b) {
    if (a == 0) return b;
    if (b == 0) return a;
    if (a == b) return a;
    if (a > b) return gcd(a - b, b);
    return gcd(a, b - a);
  }
  long long int npr(int n) {
    long long int ans = 1;
    for (int i = n; i > 1; i--) {
      ans *= i;
      if (i == n - 2) {
        ans /= 2;
      }
      ans %= 1000000007;
    }
    return ans;
  }
  long long int gcd(long long int a, long long int b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  void bfs(int n, vector<vector<int>>& edgelist, vector<int>& dist,
           queue<int>& q) {
    for (int i : edgelist[n]) {
      if (dist[i] != -1) continue;
      dist[i] = dist[n] + 1;
      q.push(i);
    }
    if (!q.empty()) {
      n = q.front();
      q.pop();
      bfs(n, edgelist, dist, q);
    }
  }
  int main() {
    long long int t;
    cin >> t;
    while (t--) {
      long long int n;
      cin >> n;
      int a[n];
      int de = 0;
      for (int i = 0; i < n; i++) {
        cin >> a[i];
      }
      bool dead = false;
      int con = 0;
      for (int i = 1; i < n; i++) {
        if (a[i] == 0 && a[i - 1] == 0) {
          dead = true;
          break;
        }
        if (a[i] == 1 && a[i - 1] == 1) {
          con++;
        }
      }
      int ans = con * 4;
      ans++;
      for (int i = 0; i < n; i++) {
        if (a[i] == 1) {
          ans += 1;
        }
      }
      if (dead) {
        cout << -1 << "\n";
      } else {
        cout << ans << "\n";
      }
    }
    return 0;
  }
  