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 MOD1 = 1000000000;
  const long long INF = 1000000000000000;
  long long vsum(vector<long long> &x) {
    long long sum = 0;
    for (auto itr1000 : x) sum += itr1000;
    return sum;
  }
  long long vsum(vector<long long> &x, long long MODULO) {
    long long sum = 0;
    for (auto itr1000 : x) {
      sum += itr1000;
      sum %= MODULO;
    };
    return sum;
  }
  long long binexpo(long long a, long long n, long long mod1000) {
    if (n <= 1) return a;
    if (n % 2) return ((a * binexpo(a, n - 1, mod1000)) % mod1000);
    long long res = (binexpo(a, n / 2, mod1000)) % mod1000;
    return (res * res) % mod1000;
  }
  long long modInverse(long long a, long long mod) {
    return binexpo(a, mod - 2, mod);
  }
  vector<long long> HAHA;
  long long factorial(long long n) {
    if (!HAHA.size()) HAHA.push_back(1);
    for (long long itr1000 = HAHA.size(); itr1000 <= n; itr1000++)
      HAHA.push_back((itr1000 * HAHA[itr1000 - 1]) % MOD1);
    return HAHA[n];
  }
  long long nCr(long long n, long long r) {
    long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
    res %= MOD1;
    res *= modInverse(factorial(r), MOD1);
    res %= MOD1;
    return res;
  }
  long long nPr(long long n, long long r) {
    long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
    res %= MOD1;
    return res;
  }
  void solve() {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      long long cnt = 0;
      long long mi = INF;
      for (long long j = i; j < n; j++) {
        if ((j - i) % 2) {
          ans += max(0ll, min(cnt, a[j]) - (cnt - min(a[i], mi)) + 1);
          if ((cnt - min(a[i], mi)) == 0) ans--;
          cnt -= a[j];
        } else {
          cnt += a[j];
        }
        mi = min(mi, cnt);
        if (mi < 0) break;
      }
      i++;
    }
    cout << ans << "\n";
  }
  bool cases = 0;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long test = 1;
    if (cases) cin >> test;
    while (test--) solve();
  }
  