description: <image>
  
  William has a favorite bracket sequence. Since his favorite sequence is quite big he provided it to you as a sequence of positive integers c_1, c_2, ..., c_n where c_i is the number of consecutive brackets "(" if i is an odd number or the number of consecutive brackets ")" if i is an even number.
  
  For example for a bracket sequence "((())()))" a corresponding sequence of numbers is [3, 2, 1, 3].
  
  You need to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.
  
  A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 1000), the size of the compressed sequence.
  
  The second line contains a sequence of integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9), the compressed sequence.
  
  Output
  
  Output a single integer — the total number of subsegments of the original bracket sequence, which are regular bracket sequences.
  
  It can be proved that the answer fits in the signed 64-bit integer data type.
  
  Examples
  
  Input
  
  
  5
  4 1 2 3 1
  
  
  Output
  
  
  5
  
  
  Input
  
  
  6
  1 3 2 1 2 4
  
  
  Output
  
  
  6
  
  
  Input
  
  
  6
  1 1 1 1 2 2
  
  
  Output
  
  
  7
  
  Note
  
  In the first example a sequence (((()(()))( is described. This bracket sequence contains 5 subsegments which form regular bracket sequences:
  
    1. Subsequence from the 3rd to 10th character: (()(()))
    2. Subsequence from the 4th to 5th character: ()
    3. Subsequence from the 4th to 9th character: ()(())
    4. Subsequence from the 6th to 9th character: (())
    5. Subsequence from the 7th to 8th character: ()
  
  
  
  In the second example a sequence ()))(()(()))) is described.
  
  In the third example a sequence ()()(()) is described.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long po(long long a, long long b, long long M) {
    long long res = 1;
    a = a % M;
    while (b) {
      if (b % 2) res = (res * a) % M;
      b /= 2;
      a = (a * a) % M;
    }
    return res;
  }
  long long pw(long long x, long long y) {
    if (y == 0)
      return 1;
    else if (y % 2 == 0)
      return po(x, y / 2, 1000000007) % 1000000007 * po(x, y / 2, 1000000007) %
             1000000007;
    else
      return x * po(x, y / 2, 1000000007) % 1000000007 *
             po(x, y / 2, 1000000007) % 1000000007;
  }
  long long modInverse(long long n, long long M) { return po(n, M - 2, M); }
  long long nCrModPFermat(long long n, long long r, long long M) {
    if (!r) return 1;
    long long fac[n + 1];
    fac[0] = 1;
    long long i;
    for (i = 1; i < n + 1; i++) {
      fac[i] = fac[i - 1] * i % M;
    }
    return (fac[n] * modInverse(fac[r], M) % M * modInverse(fac[n - r], M) % M) %
           M;
  }
  long long logk(long long a, long long b) { return log(a) / log(b); }
  bool isPrime(long long N) {
    for (long long i = 2; i * i <= N; ++i) {
      if (N % i == 0) return false;
    }
    return true;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  bool prime[1000001];
  void SieveOfEratosthenes(long long n) {
    memset(prime, true, sizeof(prime));
    for (long long p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    long long n, ans = 0;
    cin >> n;
    long long a[n + 1];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (long long i = 0; i < n; i += 2) {
      long long ominusc = a[i];
      long long open = a[i];
      for (long long j = i + 1; j < n && ominusc >= 0; j++) {
        if (j % 2) {
          ominusc -= a[j];
          ans += max(min(open, open - ominusc), 0ll);
          open = min(open, ominusc + 1);
        } else {
          ominusc += a[j];
        }
      }
    }
    cout << ans << "\n";
    return 0;
  }
  