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;
  template <typename T>
  inline bool read(T &x) {
    int f = 1, c = getchar();
    x = 0;
    while (!isdigit(c)) {
      if (c == 45) f = -1;
      c = getchar();
    }
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    return x *= f, true;
  }
  template <typename T, typename... Args>
  inline bool read(T &x, Args &...args) {
    return read(x) && read(args...);
  }
  namespace run {
  const int maxn = 1e3 + 10;
  long long a[maxn];
  long long ans, now, mini;
  int n;
  inline int main() {
    read(n);
    for (register int i = 1; i <= n; ++i) read(a[i]);
    for (register int i = 1; i <= n; i += 2) {
      now = 0, mini = 0;
      for (register int j = i + 1; j <= n; ++j) {
        if (now < mini) mini = now;
        if (!(j & 1)) {
          long long lb = -mini, rb = now + lb;
          if (lb <= a[i] && rb <= a[j]) {
            ans += min(a[i] - lb + 1, a[j] - rb + 1);
            if (!lb && !rb) --ans;
          }
          now -= a[j];
        }
        if (j & 1) now += a[j];
      }
    }
    printf("%lld\n", ans);
    return 0;
  }
  }  // namespace run
  int main() { return run ::main(); }
  