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;
  const long long inf = 1e9 + 7;
  const long long linf = 1e18 + 1;
  template <typename T>
  using V = vector<T>;
  template <typename T = long long>
  T nxt() {
    T x;
    cin >> x;
    return x;
  }
  template <typename T = long long>
  void read(vector<T>& x) {
    for (T& i : x) cin >> i;
  }
  template <typename T = long long>
  void print(vector<T>& x) {
    for (T v : x) {
      cout << v << " ";
    }
  }
  template <typename T = long long>
  void printset(set<T>& x) {
    for (auto v : x) cout << v << " ";
    cout << endl;
  }
  void print(map<char, long long>& cnt) {
    for (auto [ch, count] : cnt) {
      for (long long i = 0; i < count; ++i) {
        cout << ch;
      }
    }
  }
  long long gcd(long long a, long long b) {
    if (a == 0) {
      return b;
    }
    return gcd(b % a, a);
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long TT = 1;
    while (TT--) {
      long long n;
      cin >> n;
      vector<long long> a(n);
      read(a);
      long long ans = 0;
      for (long long l = 0; l < n; l += 2) {
        long long fp = a[l];
        long long lp = 0;
        for (long long r = l + 1; r < n; ++r) {
          if (r % 2 == 0) {
            lp += a[r];
            continue;
          }
          long long v = a[r];
          long long d = min(v, lp);
          v -= d;
          lp -= d;
          if (v == 0) {
            if (lp == 0) {
              ++ans;
            }
            continue;
          }
          if (fp != a[l]) {
            ++ans;
          }
          ans += min(fp, v);
          fp -= v;
          if (fp < 0) {
            break;
          }
        }
      }
      cout << ans << endl;
    }
    return 0;
  }
  