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_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  void _print() { cout << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    cout << t;
    if (sizeof...(v)) cout << ", ";
    _print(v...);
  }
  using ll = long long int;
  using ld = long double;
  template <typename T, typename V>
  using P = pair<T, V>;
  template <typename T>
  using V = vector<T>;
  constexpr ll MOD = 1e9 + 7;
  constexpr ll INF = 0x3f3f3f3f;
  constexpr ld PI = 3.1415926535897932384626;
  constexpr ld EPS = 1e-9;
  constexpr ll MAX = 4e6 + 5;
  void solve() {
    ll n;
    cin >> n;
    V<ll> c(n + 2, 0);
    for (ll i = (ll)(1); i < (ll)(n + 1); i++) cin >> c[i];
    ll ans = 0;
    for (ll i = (ll)(1); i < (ll)(n + 1); i++)
      if (i % 2) {
        ll cur = 0, mn = INF;
        for (ll j = (ll)(i); j < (ll)(n + 1); j++) {
          if (j % 2)
            cur += c[j];
          else {
            cur -= c[j];
            if (cur == 0)
              ans += min({mn + 1, c[i], c[j]});
            else if (cur > 0 && cur <= mn)
              ans += max(0LL, min({mn + 1 - cur, c[i] - cur, c[j]}));
            else if (cur < 0 && -cur < c[j])
              ans += max(0LL, min({mn + 1, c[i], c[j] + cur}));
          }
          if (cur < 0) break;
          mn = min(mn, cur);
        }
      }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    ll t = 1;
    for (ll i = (ll)(1); i < (ll)(t + 1); i++) solve();
  }
  