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.io.*;
  import java.util.*;
   
  public class C {
  	
      public static void main(String[] args)
      {
      	FastScanner sc=new FastScanner();
      	int t=1;
      	PrintWriter pw=new PrintWriter(System.out);
      	while(t-->0) {
      		int n=sc.nextInt();
      		int[] a=new int[n];
      		for(int i=0;i<n;i++) {
      			a[i]=sc.nextInt();
      		}
      		long ans=0;
      		for(int left=0;left<n;left+=2) {
      			long bal=0;
      			long min_bal=a[left];
      			for(int right=left+1;right<n && min_bal>=0;right+=2) {
      				bal+=a[right-1]-a[right];
      				if(bal<=min_bal) {
      					ans+=min_bal-Math.max(0, bal)+1;
      				}
      				min_bal=Math.min(min_bal, bal);
      			}
      		}
      		pw.println(ans-n/2);
      	}
      	
      	pw.flush();
      }
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  
  }