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.*;
  import static java.lang.Math.*;
  
  public class C{
  	public static void main(String[] args) throws IOException {
  		// br = new BufferedReader(new FileReader(".in"));
  		// out = new PrintWriter(new FileWriter(".out"));
  		//new Thread(null, new (), "peepee", 1<<28).start();
  		
  		long ans = 0;
  		int n =readInt();
  		long[] a = new long[n];
  		for (int i = 0; i  <n ;i++) a[i]=readInt();
  		
  		long[][] left = new long[n][n];
  		long[][] right =new long[n][n];
  		
  		// Got my dfirections mixed up
  		// Left is (, right is ).
  		
  		for (int i = 0; i <n; i++) {
  			for (int j = i; j < n;j++) {
  				if (i==j) {
  					if (i%2==0) left[i][j] = a[i];
  					else right[i][j] = a[i];
  				}
  				else {
  					left[i][j] += left[i][j-1];
  					right[i][j] += right[i][j-1];
  					if (j%2==0) {
  						// Left facing can never cancel out right facing
  						left[i][j] += a[j];
  					}
  					else {
  						// Right facing cancels out left facing
  						long sub = min(left[i][j], a[j]);
  						left[i][j] -= sub;
  						right[i][j] += a[j]-sub;
  					}
  				}
  			}
  		}
  		
  		//System.out.println(left[1][2] + " " + right[1][2]);
  		
  		for (int i = 0; i < n; i++) {
  			for (int j = i+1 ; j < n; j++) {
  				if (i+1==j) {
  					if (i%2==0) {
  					//	System.out.println("on " + i + " " + j + ", cheeky add " + min(a[i],a[j]));
  						ans += min(a[i],a[j]);
  					}
  					continue;
  				}
  				if (i%2!=0 || j%2!=1) continue;
  				// Okay, so now you have the middle segment.
  				// Now do some big brain shit/
  				// Cancel out first, then add and gg
  				long tempR =  a[j] - left[i+1][j-1];
  				long tempL =  a[i] - right[i+1][j-1];
  				if (tempR < 0||tempL < 0) continue;
  				long add = min(tempR, tempL)+1;
  				//System.out.println(a[i] + " " + a[j]);
  				//System.out.println("Hello " + left[i+1][j-1] + " " + right[i+1][j-1] + " ");
  				//System.out.println("Yo " + i + " " + j + " " + tempL + " " + tempR + " added " +add);
  				ans += add;
  			}
  		}
  		out.println(ans);
  		
  		out.close();
  	}
  	
  	/* Stupid things to try if stuck:
  	 * n=1, expand bsearch range
  	 * brute force small patterns
  	 * submit stupid intuition
  	 */
  	
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static StringTokenizer st = new StringTokenizer("");
  	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
  	static int readInt() throws IOException{return Integer.parseInt(read());}
  	static long readLong() throws IOException{return Long.parseLong(read());}
  	static double readDouble() throws IOException{return Double.parseDouble(read());}
  }