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.io.*;
  public class C {
  
  	public static void main(String[] args) {
  		FastScanner sc = new FastScanner();
  		int n = sc.nextInt();
  		long[] arr = new long[n];
  		for(int i = 0; i < n; i++) {
  			arr[i] = sc.nextLong();
  		}
  		if(n % 2 == 1) n--;
  		long[] brr = new long[n+1];
  		brr[0] = 0;
  		TreeSet<Long> vals = new TreeSet<>();
  		vals.add(0L);
  		for(int i = 1; i <= n; i++) {
  			if(i % 2 == 1) {
  				brr[i] = brr[i-1] + arr[i-1];
  			}
  			else {
  				brr[i] = brr[i-1] - arr[i-1];
  			}
  			vals.add(brr[i]);
  		}
  		long res = 0;
  		for(long h: vals) {
  			long count = 1;
  			boolean start = false;
  			for(int i = 0; i < n; i += 2) {
  				if(brr[i] < h && brr[i+1] > h) {
  					if(start) {
  						res += count; count = 1;
  					}
  					start = true;
  				}
  				else if(brr[i] == h) {
  					if(start) {
  						res += count; count++;
  					}
  					start = true;
  				}
  				else {
  					//nothing
  				}
  			}
  			if(start && brr[n] <= h) {
  				res += count;
  			}
  		}
  		long gaps = 0;
  		long prevh = Long.MIN_VALUE;
  		for(long h: vals) {
  			res += gaps * (h - prevh - 1);
  			gaps = 0;
  			for(int i = 0; i < n; i += 2) {
  				if(brr[i] <= h && brr[i+1] > h) gaps++;
  			}
  			if(brr[n] > h) gaps--;
  			prevh = h;
  		}
  		System.out.println(res);
  	}
  	
  	static class FastScanner {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  		public FastScanner() {
  			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
  			tokenizer = null;
  		}
  		public String next() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch(IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  	}
  
  }
  