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 javax.swing.plaf.synth.SynthSpinnerUI;
  
  import java.io.*;
  import java.net.StandardSocketOptions;
  
  public class tr0 {
  	static PrintWriter out;
  	static StringBuilder sb;
  	static long mod = (long) 1e9 + 7;
  	static long inf = (long) 1e16;
  	static int n, k;
  	static ArrayList<Integer>[] ad;
  	static int[][] remove, add;
  	static int[][] memo;
  	static boolean vis[];
  	static long[] inv, ncr[];
  	static HashMap<Integer, Integer> hm;
  	static int[] pre, suf, Smax[], Smin[];
  	static int idmax, idmin;
  	static ArrayList<Integer> av;
  	static HashMap<Integer, Integer> mm;
  	static boolean[] msks;
  	static int[] lazy[], lazyCount;
  	static int[] dist;
  	static int[][] P;
  	static long N;
  	static ArrayList<Integer> gl;
  	static int[] a;
  	static String s;
  
  	public static void main(String[] args) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		out = new PrintWriter(System.out);
  		int n = sc.nextInt();
  		int[] a = sc.nextArrInt(n);
  		long ans = 0;
  		long[][] maxneg = new long[n][n];
  		for (int i = 0; i < n; i++) {
  			if (i % 2 == 1)
  				continue;
  			long sum = 0;
  			long min = Long.MAX_VALUE;
  			for (int j = i + 1; j < n; j++) {
  				if (j % 2 == 0)
  					sum += a[j];
  				else
  					sum -= a[j];
  				min = Math.min(min, sum);
  				maxneg[i][j] = min;
  			}
  		}
  		for (int i = 0; i < n; i++) {
  			if (i % 2 == 1) {
  				ans += Math.min(a[i], a[i - 1]);
  				long sum = -a[i] + a[i - 1];
  				long min = 0;
  				for (int j = i - 3; j >= 0; j -= 2) {
  					if (sum > 0)
  						break;
  					sum += -a[j + 1];
  					min = maxneg[j][i - 1];
  					ans += Math.max(0, Math.min(a[j], -sum) + min + 1);
  					sum += a[j];
  				}
  			}
  		}
  		System.out.println(ans);
  		out.flush();
  	}
  
  	static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  
  		public Long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public int[] nextArrInt(int n) throws IOException {
  			int[] a = new int[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextInt();
  			return a;
  		}
  
  		public long[] nextArrLong(int n) throws IOException {
  			long[] a = new long[n];
  			for (int i = 0; i < n; i++)
  				a[i] = nextLong();
  			return a;
  		}
  
  		public int[] nextArrIntSorted(int n) throws IOException {
  			int[] a = new int[n];
  			Integer[] a1 = new Integer[n];
  			for (int i = 0; i < n; i++)
  				a1[i] = nextInt();
  			Arrays.sort(a1);
  			for (int i = 0; i < n; i++)
  				a[i] = a1[i].intValue();
  			return a;
  		}
  
  		public long[] nextArrLongSorted(int n) throws IOException {
  			long[] a = new long[n];
  			Long[] a1 = new Long[n];
  			for (int i = 0; i < n; i++)
  				a1[i] = nextLong();
  			Arrays.sort(a1);
  			for (int i = 0; i < n; i++)
  				a[i] = a1[i].longValue();
  			return a;
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  	}
  }