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.util.Map.Entry;
  
  import java.math.*;
  import java.io.*;
  
  public class Main {
  	public static void main(String[] args) throws FileNotFoundException {
  		InputReader in = new InputReader(System.in);
  		// Scanner in = new Scanner(System.in);
  		// Scanner in = new Scanner(new BufferedReader(new
  		// InputStreamReader(System.in)));
  		PrintWriter out = new PrintWriter(System.out);
  		// InputReader in = new InputReader(new
  		// File("ethan_traverses_a_tree.txt"));
  		// PrintWriter out = new PrintWriter(new
  		// File("ethan_traverses_a_tree-output.txt"));
  
  		int n = in.nextInt();
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++) {
  			a[i] = in.nextInt();
  		}
  
  		long ans = 0;
  		for (int i = 0; i < n; i = i + 2) {
  			long sum = 0;
  			long min = 0;
  			for (int j = i + 1; j < n; j++) {
  				if (j == i + 1) {
  					ans = ans + Math.min(a[i], a[j]);
  				} else if (j % 2 == 1) {
  					long substract1 = -min;
  					if (a[i] - substract1 >= 0) {
  						long k1 = a[i] - substract1;
  						long nowsum = sum + substract1;
  
  						if (a[j] >= nowsum) {
  							long k2 = a[j] - nowsum;
  
  							ans = ans + 1 + Math.min(k1, k2);
  						}
  					}
  				}
  
  				if (j % 2 == 0) {
  					sum = sum + a[j];
  				} else {
  					sum = sum - a[j];
  				}
  
  				min = Math.min(min, sum);
  			}
  		}
  		out.printf("%d\n", ans);
  
  		out.close();
  	}
  
  	static class InputReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public InputReader(File f) {
  			try {
  				br = new BufferedReader(new FileReader(f));
  			} catch (FileNotFoundException e) {
  				e.printStackTrace();
  			}
  		}
  
  		public InputReader(InputStream in) {
  			br = new BufferedReader(new InputStreamReader(in));
  		}
  
  		public String next() {
  			while (st == null || !st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		public boolean hasNext() {
  			while (st == null || !st.hasMoreTokens()) {
  				String s = null;
  				try {
  					s = br.readLine();
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  				if (s == null)
  					return false;
  				st = new StringTokenizer(s);
  			}
  			return true;
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }