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: // Problem: Compressed Bracket Sequence
  // https://codeforces.com/contest/1556/problem/C
  
  import java.util.*;
  
  public class CompressedBracketSequence {
  
    private static class BracketPair {
      private long open, close;
    }
  
    // Cache for the number of valid solutions.
    // The element `cache.get(i)` stores the number of valid solutions using sequence `v[i, ]`
    // with `key` additonal open brackets at the beginning.
    private static ArrayList<Map<Long, Long>> cache;
  
    private static ArrayList<BracketPair> v;
  
    private static long f(int i, long open) {
      if (i >= v.size()) {
        return 0;
      }
  
      if (cache.get(i).containsKey(open)) {
        return cache.get(i).get(open);
      }
  
      BracketPair p = v.get(i);
      long result = 0;
      if (open + p.open < p.close) {
        result = open + p.open;
      }  else if (p.close >= 1) {
        // open + p.open >= p.close >= 1
        result = (p.close - 1) + 1 + g(i + 1);
      }
  
      result += f(i + 1, Math.max(0, open + p.open - p.close));
  
      cache.get(i).put(open, result);
      return result;
    }
  
    private static Long[] g;
  
    private static long g(int i) {
      if (i >= v.size()) {
        return 0;
      }
  
      if (g[i] != null) {
        return g[i];
      }
  
      BracketPair p = v.get(i);
      long result = 0;
      if (p.open < p.close) {
        result = 1;
      } else if (p.open == p.close) {
        result = 1 + g(i + 1);
      } else {
        // p.open > p.close
        long open = p.open - p.close;
        for (int j = i + 1; j < v.size(); ++j) {
          open += v.get(j).open;
          if (open < v.get(j).close) {
            result = 1;
            break;
          } else if (open > v.get(j).close) {
            open -= v.get(j).close;
          } else {
            // open == v.get(j).close
            result = 1 + g(j + 1);
            break;
          }
        }
      }
    
      return g[i] = result;
    }
  
    public static void main(String[] args) {
      Scanner in = new Scanner(System.in);
  
      int n = in.nextInt();
      v = new ArrayList<>();
      BracketPair current = new BracketPair();
  
      for (int i = 0; i < n; ++i) {
        int x = in.nextInt();
  
        if (i % 2 == 0) {
          // Open brackets.
          if (current.close > 0) {
            v.add(current);
            current = new BracketPair();
          }
  
          current.open += x;
        } else {
          // Close brackets.
          current.close += x;
        }
      }
      v.add(current);
  
      cache = new ArrayList<>();
      for (int i = 0; i < v.size(); ++i) {
        cache.add(new HashMap<>());
      }
  
      g = new Long[v.size()];
  
      System.out.println(f(0, 0));
  
      in.close();
    }
  }