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;
  const signed long long MOD = 1e9 + 7;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  signed long long n;
  signed long long c[1001];
  pair<signed long long, signed long long> intersect(
      pair<signed long long, signed long long> a,
      pair<signed long long, signed long long> b) {
    if (a.first > b.second || b.first > a.second) return {1, 0};
    return {max(a.first, b.first), min(a.second, b.second)};
  }
  signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n;
    for (signed long long i = 1; i <= n; i++) cin >> c[i];
    signed long long ans = 0;
    signed long long open;
    signed long long close = 0;
    for (signed long long i = 1; i <= n; i += 2) {
      open = close = 0;
      signed long long ok = 1;
      for (signed long long j = i + 1; j <= n; j++) {
        if (j & 1)
          open += c[j];
        else {
          signed long long d = open - close;
          pair<signed long long, signed long long> use =
              intersect({1, c[j]}, {d + ok, d + c[i]});
          if (use.second >= use.first) ans += use.second - use.first + 1;
          close += c[j];
          ok = max(ok, close - open);
          if (ok > c[i]) break;
        }
      }
    }
    cout << ans << "\n";
  }
  