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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")
  #pragma comment(linker, "/STACK:64777216")
  using namespace std;
  const int INF = int(1e9);
  const long long INF64 = long long(1e18);
  const long double PI = acosl(long double(-1));
  const long double EPS = 1e-9;
  template <typename T>
  inline T sqr(const T& first) {
    return first * first;
  }
  template <typename T>
  inline T abs(const T& first) {
    return first > 0 ? first : -first;
  }
  inline bool inside(int first, int second, int n, int m) {
    return first >= 1 && first <= n && second >= 1 && second <= m;
  }
  inline int rnd() { return abs(rand() ^ (rand() << 15)); }
  inline int rnd(int n) {
    assert(n > 0);
    return rnd() % n;
  }
  inline int rnd(int lf, int rg) { return lf + rnd(rg - lf + 1); }
  inline long long rndLL() { return rnd() * 1LL * rnd() + rnd(); }
  const int dx[4] = {-1, 0, +1, 0};
  const int dy[4] = {0, +1, 0, -1};
  const int dx8[8] = {-1, -1, 0, +1, +1, +1, 0, -1};
  const int dy8[8] = {0, +1, +1, +1, 0, -1, -1, -1};
  const int N = int(2e5) + 555;
  int n;
  long long a[N];
  void gen() {}
  bool read() {
    if (!(cin >> n)) return false;
    for (int i = 1; i <= int(n); i++) assert(cin >> a[i]);
    return true;
  }
  long long inter(long long a1, long long b1, long long a2, long long b2) {
    long long lf = max(a1, a2);
    long long rg = min(b1, b2);
    return max(0LL, rg - lf + 1);
  }
  void solve() {
    long long ans = 0;
    for (int i = 1; i + 1 <= n; i += 2) {
      ans += min(a[i], a[i + 1]);
    }
    for (int i = 1; i + 3 <= n; i += 2) {
      long long bal = 0;
      long long mn = a[i + 1];
      bal += a[i + 2];
      if (mn > a[i]) continue;
      for (int j = i + 3; j <= n; j += 2) {
        ans += inter(bal, bal + (a[i] - mn), 1, a[j]);
        bal -= a[j];
        if (bal < 0) {
          mn += -bal;
          bal = 0;
        }
        if (mn > a[i]) break;
        bal += a[j + 1];
      }
    }
    cout << ans << endl;
  }
  long double gett() { return long double(clock()) / CLOCKS_PER_SEC; }
  int main() {
    cout << setprecision(15) << fixed;
    cerr << setprecision(15) << fixed;
    int T = 1;
    srand(int(time(NULL)));
    for (int i = 0; i < int(T); i++) {
      assert(read());
      solve();
    }
    return 0;
  }
  