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;
  long long gcd(long long a, long long b) {
    if (a < 0) a = -a;
    if (b < 0) b = -b;
    return b == 0 ? a : gcd(b, a % b);
  }
  template <typename T>
  void read(T &res) {
    bool flag = false;
    char ch;
    while (!isdigit(ch = getchar())) (ch == '-') && (flag = true);
    for (res = ch - 48; isdigit(ch = getchar());
         res = (res << 1) + (res << 3) + ch - 48)
      ;
    flag && (res = -res);
  }
  long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
  long long qp(long long a, long long b, long long mod) {
    long long ans = 1;
    if (b == 0) {
      return ans % mod;
    }
    while (b) {
      if (b % 2 == 1) {
        b--;
        ans = ans * a % mod;
      }
      a = a * a % mod;
      b = b / 2;
    }
    return ans % mod;
  }
  long long qpn(long long a, long long b, long long p) {
    long long ans = 1;
    a %= p;
    while (b) {
      if (b & 1) {
        ans = (ans * a) % p;
        --b;
      }
      a = (a * a) % p;
      b >>= 1;
    }
    return ans % p;
  }
  signed main() {
    long long t;
    read(t);
    while (t--) {
      long long n;
      read(n);
      long long f = 0;
      long long h = 1;
      long long op = 0;
      long long pre = 0;
      for (long long i = 1; i <= n; i++) {
        long long x;
        read(x);
        if (x == 1) {
          if (pre == 1) {
            h = h + 5;
          } else {
            h++;
          }
          pre = x;
          f = 0;
          continue;
        }
        f++;
        pre = 0;
        if (f == 2) {
          op = 1;
        }
      }
      if (op) {
        printf("-1\n");
        continue;
      }
      printf("%lld\n", h);
    }
  }
  