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;
  long long n, a[1010];
  long long res[1010 * 1010];
  long long bf() {
    int m = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < a[i]; j++) res[m++] = ((i % 2) ? -1 : 1);
    }
    long long ans = 0;
    for (int i = 0; i < m; i++) {
      int cv = res[i];
      if (cv < 0) continue;
      for (int j = i + 1; j < m; j++) {
        cv += res[j];
        if (cv < 0) break;
        if (cv == 0) ans++;
      }
    }
    return ans;
  }
  long long solve() {
    long long ans = 0;
    for (long long i = 0; i < n; i += 2) {
      long long between = 0;
      long long minbetween = 0;
      for (long long j = i + 1; j < n; j += 2) {
        long long rem1 = a[i], rem2 = a[j];
        rem1 += minbetween;
        long long cb = between - minbetween;
        rem2 -= cb;
        if (rem1 >= 0 && rem2 >= 0) {
          ans += min(rem1, rem2);
          if (minbetween != 0) ans++;
        }
        between -= a[j];
        minbetween = min(minbetween, between);
        if (j + 1 < n) between += a[j + 1];
      }
    }
    return ans;
  }
  void gen(int c, int mv, int fc) {
    if (c < 0) {
      if (solve() != bf()) {
        cout << solve() << " " << bf() << endl;
        for (int i = 0; i <= fc; i++) cout << a[i] << " ";
        cout << endl;
      }
      return;
    }
    for (int i = 1; i <= mv; i++) {
      a[c] = i;
      gen(c - 1, mv, fc);
    }
  }
  int main() {
    scanf("%lld", &n);
    for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
    printf("%lld\n", solve());
  }
  