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;
  clock_t startTime, endTime;
  const long long INF_ll = 1e18;
  const int INF_int = 0x3f3f3f3f;
  void read(){};
  template <typename _Tp, typename... _Tps>
  void read(_Tp& x, _Tps&... Ar) {
    x = 0;
    char c = getchar();
    bool flag = 0;
    while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
    while (c >= '0' && c <= '9')
      x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (flag) x = -x;
    read(Ar...);
  }
  template <typename T>
  inline void write(T x) {
    if (x < 0) {
      x = ~(x - 1);
      putchar('-');
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
  }
  void rd_test() {}
  void Time_test() {}
  int vec[300000];
  int main() {
    int n;
    read(n);
    for (int i = 1; i <= n; i++) {
      read(vec[i]);
    }
    long long ans = 0;
    int len = n;
    if (len % 2 == 1) len--;
    for (int i = 1; i <= len; i += 2) {
      long long sum = vec[i];
      long long L = 0;
      for (int j = i + 1; j <= len; j++) {
        if (j % 2 == 1)
          L += vec[j];
        else {
          long long R = vec[j];
          long long k = min(L, R);
          if (k) {
            R -= k;
            L -= k;
            if (L == 0) ans++;
          }
          if (R > 0) {
            long long k = min(R, sum);
            R -= k;
            sum -= k;
            ans += k;
          }
          if (R > 0) break;
        }
      }
    }
    printf("%lld\n", ans);
  }
  