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;
  long long n;
  long long a[500505];
  signed main() {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i += 2) {
      long long s = 1;
      long long rm = a[i] - 1;
      for (long long j = i + 1; j <= n; j++) {
        if (j % 2 == 0) {
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          s += 1;
          s -= 1;
          if (s == 0)
            ans += min(rm + 1, a[j]);
          else if (s < 0) {
            long long z = rm + s;
            if (z < 0) break;
            s = 0;
            ans += min(z + 1, a[j]);
            ;
          } else {
            if (s <= a[j] - 1) ans += min(a[j] - s, rm + 1);
          };
          s -= a[j];
          s++;
          if (s < 0) {
            rm += s;
            s = 0;
          }
          if (rm + 1 < 1) break;
        } else
          s += a[j];
      }
    }
    cout << ans;
  }
  