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;
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long T = 1;
    while (T--) {
      long long n;
      cin >> n;
      vector<long long> v(n);
      for (long long i = 0; i < n; i++) cin >> v[i];
      long long count = 0;
      for (long long i = 0; i < n; i += 2) {
        long long cur = 0;
        long long diff = 0;
        for (long long j = i + 1; j < n; j++) {
          if ((j - i) % 2 == 0) {
            cur += v[j];
            continue;
          }
          long long p1, p2;
          p1 = max(-diff, 1ll);
          p2 = v[i];
          if (p1 > p2) break;
          long long aa = p1 + cur, bb = p2 + cur;
          if (max(1ll, aa) <= min(v[j], bb)) {
            count += min(v[j], bb) - max(1ll, aa) + 1;
          }
          cur -= v[j];
          diff = min(diff, cur);
        }
      }
      cout << count << "\n";
    }
  }
  