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: import java.util.*;
  import java.io.*;
  
  public class Deltix {
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = 1;
          while (t-- > 0) {
              int n = sc.nextInt();
              long [] c = new long[n];
              for (int i = 0; i < n; i++) c[i] = sc.nextLong();
              long res = 0;
              for (int i = 0; i < n; i++) {
                  if (i % 2 == 1) continue;
                  long in = 0;
                  long low = 1;
                  // impose at least 1 from each
                  for (int j = i + 1; j < n; j++) {
                      if (j % 2 == 1) {
                          // in positive means more left
                          if (in == 0) {
                              res += Math.max(0, Math.min(c[i], c[j]) - low + 1);
                          } else {
                              res += ans(c[i], c[j], -in, low);
                          }
                      }
                      if (j % 2 == 1) {
                          low = Math.max(low, c[j] - in);
                          in -= c[j];
                      } else {
                          in += c[j];
                      }
                  }
              }
              out.println(res);
          }
          out.close();
      }
  
      static long ans(long a, long b, long dif, long minA) {
          if (minA > a) return 0;
          if (dif > 0) {
              // a ranges from minA to a
              // b ranges from 1 to b
              // want to create dif of dif
              long start = Math.max(minA, dif + 1);
              long end = Math.min(a, b + dif);
              return Math.max(0, end - start + 1);
          } else {
              dif = -dif;
              long start = minA;
              long end = Math.min(a, b - dif);
              return Math.max(0, end - start + 1);
          }
      }
  
  
  
  
      static void sort(int[] a) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> q = new ArrayList<>();
          for (long i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
  
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
  
      }
  
  }