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 Test {
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static StringTokenizer st;
  	static PrintWriter pr = new PrintWriter(new OutputStreamWriter(System.out));
  	static String readLine() throws IOException {
  		return br.readLine();
  	}
  	static String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(readLine());
  		return st.nextToken();
  	}
  	static int readInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  	static long readLong() throws IOException {
  		return Long.parseLong(next());
  	}
  	static double readDouble() throws IOException {
  		return Double.parseDouble(next());
  	}
  	static char readChar() throws IOException {
  		return next().charAt(0);
  	}
  	static class Pair implements Comparable<Pair> {
  		int f, s;
  		Pair(int f, int s) {
  			this.f = f; this.s = s;
  		}
  		public int compareTo(Pair other) {
  			if (this.f != other.f) return this.f - other.f;
  			return this.s - other.s;
  		}
  	}
  	final static long inf = (long)1e18;
  	static void solve() throws IOException {
  		int n = readInt();
  		long ans = 0, c[] = new long[n + 1];
  		for (int i = 1; i <= n; ++i) {
  			c[i] = readInt();
  			if ((i&1) == 1) continue; 
  			c[i] *= -1;
  			long sum = 0, min = inf;
  			for (int j = i - 1; j >= 1; --j) {	
  				if ((j&1) == 1) {
  					long lo = Math.max(-sum, -min), hi = Math.min(c[j], -c[i] - sum);
  					if (sum + lo == 0) ++lo;
  					if (lo <= hi) ans += hi - lo + 1;
  				}
  				sum += c[j];
  				min = Math.min(min + c[j], c[j]);
  			}
  		}
  		pr.println(ans);
  	}
  	public static void main(String[] args) throws IOException {
  		solve();
  		//for (int t = readInt(); t > 0; --t) solve();
  		pr.close();
  	}
  }
  