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 static java.lang.Integer.parseInt;
  import static java.lang.Long.parseLong;
  import static java.lang.Math.min;
  import static java.lang.System.exit;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class C {
  
  	static void solve() throws Exception {
  		int n = scanInt();
  		long depth = 0, ans = 0, mvals[] = new long[n];
  		int mcnt = 0, mcnts[] = new int[n];
  		for (int i = 0; i < n; i++) {
  			int c = scanInt();
  			if ((i & 1) == 0) {
  				depth += c;
  			} else {
  				ans += min(depth, c);
  				depth -= c;
  				while (mcnt > 0 && mvals[mcnt - 1] > depth) {
  					ans += mcnts[mcnt - 1];
  					--mcnt;
  				}
  				if (mcnt > 0 && mvals[mcnt - 1] == depth) {
  					ans += mcnts[mcnt - 1];
  					++mcnts[mcnt - 1];
  				} else if (depth < 0) {
  					depth = 0;
  				} else {
  					mvals[mcnt] = depth;
  					mcnts[mcnt] = 1;
  					++mcnt;
  				}
  			}
  		}
  		out.print(ans);
  	}
  
  	static int scanInt() throws IOException {
  		return parseInt(scanString());
  	}
  
  	static long scanLong() throws IOException {
  		return parseLong(scanString());
  	}
  
  	static String scanString() throws IOException {
  		while (tok == null || !tok.hasMoreTokens()) {
  			tok = new StringTokenizer(in.readLine());
  		}
  		return tok.nextToken();
  	}
  
  	static BufferedReader in;
  	static PrintWriter out;
  	static StringTokenizer tok;
  
  	public static void main(String[] args) {
  		try {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(System.out);
  			solve();
  			in.close();
  			out.close();
  		} catch (Throwable e) {
  			e.printStackTrace();
  			exit(1);
  		}
  	}
  }