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 long long INF = 1e18;
  const int N = 3e5 + 5;
  int rand(int a, int b) { return a + rand() % (b - a + 1); }
  int brute(vector<int>& c) {
    int n = c.size();
    string s;
    for (int i = 0; i < n; ++i) {
      int x = c[i];
      if (i % 2) {
        while (x--) {
          s.push_back(')');
        }
      } else {
        while (x--) {
          s.push_back('(');
        }
      }
    }
    int cnt = 0;
    n = s.size();
    for (int i = 0; i < n; ++i) {
      int sum = 0;
      for (int j = i; j < n; ++j) {
        if (s[j] == '(') {
          sum++;
        } else {
          sum--;
        }
        if (sum < 0) {
          break;
        }
        cnt += (sum == 0);
      }
    }
    return cnt;
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> c(n);
    for (auto& it : c) {
      cin >> it;
    }
    long long answer = 0;
    for (int i = 0; i < n; i += 2) {
      long long curr = c[i];
      long long mn = INF;
      for (int j = i + 1; j < n; ++j) {
        if (j % 2 == 0) {
          curr += c[j];
        } else {
          curr -= c[j];
          if (curr < 0 && -curr < c[j]) {
            long long R = mn + 1;
            long long L = 1;
            answer += max(0LL, min(R - L + 1, c[j] + curr));
          }
          if (curr > 0 && curr <= mn) {
            long long R = mn + 1;
            long long L = curr + 1;
            answer += max(0LL, min(R - L + 1, 1LL * c[j]));
          }
          if (curr == 0) {
            answer += min({mn + 1, 1ll * c[i], 1ll * c[j]});
          }
        }
        mn = min(mn, curr);
        if (curr < 0) {
          break;
        }
      }
    }
    cout << answer << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    int i = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  