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.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class taskc {
  	public static void main(String[] args) {
  		FastScanner in = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int n = in.nextInt();
  		long a[] = new long[n];
  		for (int i = 0; i < n; i++) {
  			a[i] = in.nextLong();
  		}
  		long b[][] = new long[n][n];
  		long mn[][] = new long[n][n];
  		for (int i = 0; i < n; i++) {
  			long sum = a[i] * (i % 2 == 0 ? 1L : -1L);
  			long min = sum;
  			for (int j = i + 1; j < n; j++) {
  				sum += a[j] * (j % 2 == 0 ? 1L : -1L);
  				min = Math.min(min, sum);
  				b[i][j] = sum;
  				mn[i][j] = min;
  			}
  		}
  		long ans = 0;
  		for (int i = 0; i < n; i++) {
  			for (int j = i + 1; j < n; j++) {
  				if (i % 2 == 0 && j % 2 == 1) {
  					long mid = (i + 1 < j - 1 ? b[i + 1][j - 1] : 0L);
  					long mid_min = (i + 1 < j - 1 ? mn[i + 1][j - 1] : 0L);
  					// use x from left to make x + mid_min >= 0 && x + mid - y = 0;
  					long from_x = Math.max(1, -mid_min), to_x = a[i];
  					if (from_x > to_x) {
  						continue;
  					}
  					long from_y = Math.max(1, from_x + mid);
  					long to_y = Math.min(a[j], to_x + mid);
  					if (from_y <= to_y) {
  						ans += to_y - from_y + 1;
  					}
  				}
  			}
  		}
  		out.println(ans);
  		out.flush();
  	}
  
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer tok = new StringTokenizer("");
  		String next() {
  			while(!tok.hasMoreTokens()) {
  				try {
  					tok = new StringTokenizer(br.readLine());
  				} catch (Exception e) {
  					e.printStackTrace();
  				}
  			}
  			return tok.nextToken();
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		float nextFloat() {
  			return Float.parseFloat(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  }
  