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.lang.*;
  import java.io.*;
  
  public class Main
  {
  	PrintWriter out = new PrintWriter(System.out);
  	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer tok = new StringTokenizer("");
      String next() throws IOException {
          if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
          return tok.nextToken();
      }
      int ni() throws IOException { return Integer.parseInt(next()); }
      long nl() throws IOException { return Long.parseLong(next()); }
      
      long mod=1000000007;
      
      void solve() throws IOException {
          int n=ni();
          long ct=0;
          long ans=0;
          long currmin=0;
          Stack<Long>S=new Stack();
          Stack<Long>T=new Stack();
          S.push(0L);
          T.push(0L);
          
          for (int i=0;i<n;i++) {
              int u=ni();
              if (i%2==0) ct+=u;
              else {
                  ans+=Math.min(ct-currmin,u);
                  ct-=u;
                  //if (ct<0) ct=0;
                  while (!S.empty() && ct<S.peek()) {
                      S.pop();
                      ans+=T.pop();
                  }
                  if (S.empty()) {
                      S.push(ct);
                      T.push(0L);
                      currmin=ct;
                  }
                  else {
                      if (ct==S.peek()) {
                          long v=T.pop();
                          ans+=v;
                          T.push(v+1);
                      }
                      else {
                          S.push(ct);
                          T.push(1L);
                      }
                  }
              }
          }
          out.println(ans);
          out.flush();
      }
      
      int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
      long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
      long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
      
      public static void main(String[] args) throws IOException {
          new Main().solve();
      }
  }