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 DeltixSU21C {
  	public static void solve(IO io) {
  
  		int k = io.getInt();
  		
  		long [] arr = new long[k];
  		for (int i = 0; i < k; i++) {
  			arr[i] = io.getLong();
  			
  		}
  		long total = 0;
  		for (int i = 0; i < k; i+= 2) {
  
  			long hl = 1;
  			long hh = arr[i];
  			
  			for (int j = i + 1; j < k; j++) {
  				
  				if (j % 2 == 0) {
  					hl += arr[j];
  					hh += arr[j];
  				} else {
  					hl -= arr[j];
  					hh -= arr[j];
  					
  					if (hh < 0) {
  						total+= (hh - hl + 1);
  						break;
  					} else if (hl <= 0) {
  						total += 0 - hl + 1;
  						hl = 0;
  					}
  
  				}
  			}
  		}
  		
  		io.println(total);
  	}
  
  	public static void main(String[] args) {
  		IO io = new IO();
  		solve(io);
  		io.close();
  	}
  
  	static class IO extends PrintWriter {
  
  		public IO() {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		public IO(InputStream i) {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public IO(InputStream i, OutputStream o) {
  			super(new BufferedOutputStream(o));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public boolean hasMoreTokens() {
  			return peekToken() != null;
  		}
  
  		public int getInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double getDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long getLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public String getWord() {
  			return nextToken();
  		}
  
  		private BufferedReader r;
  		private String line;
  		private StringTokenizer st;
  		private String token;
  
  		private String peekToken() {
  			if (token == null)
  				try {
  					while (st == null || !st.hasMoreTokens()) {
  						line = r.readLine();
  						if (line == null)
  							return null;
  						st = new StringTokenizer(line);
  					}
  					token = st.nextToken();
  				} catch (IOException e) {
  				}
  			return token;
  		}
  
  		private String nextToken() {
  			String ans = peekToken();
  			token = null;
  			return ans;
  		}
  	}
  }
  