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;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  long long INF = 1e18 + 7;
  long long MOD = 998244353;
  template <typename T>
  T my_gcd(T a, T b) {
    return gcd(a, b);
  }
  template <typename T>
  T my_min(T a, T b) {
    return min(a, b);
  }
  template <typename T>
  T my_max(T a, T b) {
    return max(a, b);
  }
  template <typename T>
  T my_and(T a, T b) {
    return (a & b);
  }
  template <typename T>
  T my_xor(T a, T b) {
    return (a ^ b);
  }
  template <typename T>
  T my_or(T a, T b) {
    return (a | b);
  }
  template <typename T>
  T my_sum(T a, T b) {
    return (a + b);
  }
  template <typename T>
  T my_sum_mod(T a, T b) {
    return (a + b) % MOD;
  }
  template <typename T>
  T my_mod(T a, T mod) {
    return a % mod;
  }
  int solve() {
    long long n;
    while (!(cin >> n)) {
      return 1;
    }
    vector<long long> a(n + 1);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    for (int i = 0; i < n - 1; i += 2) {
      long long diff = 0, mn = INF;
      for (int j = i + 1; j < n; j += 2) {
        ans += max(0ll, min(1ll * a[i], a[j] - diff) + min(mn, diff - 1) + 1);
        mn = min(mn, diff - a[j]);
        diff += (-a[j] + a[j + 1]);
      }
    }
    cout << ans << endl;
    return 0;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cout << std::fixed << std::setprecision(15);
    cin.tie(0);
    int TET = 1e9;
    for (int i = 1; i <= TET; i++) {
      if (solve()) {
        break;
      }
    }
  }
  