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: /*==========================================================================
  *   	AUTHOR:    RonWonWon
  *   	CREATED:   03.09.2021 11:21:20        
  /*==========================================================================*/
  import java.io.*;
  import java.util.*;
  
  public class C {
  	
  	public static void main(String[] args) throws IOException {
  		int n = in.nextInt();
  		int a[] = in.readArray(n);
  		if(n%2==1) n--;
  		long ans = 0;
  		for(int i=0;i<n;i+=2){
  			ans += Math.min(a[i],a[i+1]);
  			long sum = a[i]-a[i+1], neg = -sum;
  			int j = i-2;
  			//out.println(i+" "+ans);
  			while(j>-1 && sum<=0){
  				int d = a[j] - a[j+1];
  				sum += d;
  				if(d>=0){
  					if(sum<=0){
  						if(-sum<=neg){
  							ans += Math.min(d+1,neg+sum+1);
  							neg = -sum;
  						}
  					}
  					else{
  						ans += Math.min(d+1,neg+1);
  					}
  				}
  				j-=2;
  			}
  			//out.println(i+" "+ans);
  		}
  		out.println(ans);
      out.flush();
  	}
  	
  	static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
  	static int oo = Integer.MAX_VALUE;
  	static long ooo = Long.MAX_VALUE;
  	
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		void ini() throws FileNotFoundException {
  			br = new BufferedReader(new FileReader("input.txt"));
  		}
  		
  		String next() {
  			while(!st.hasMoreTokens())
  				try { st = new StringTokenizer(br.readLine()); }
  				catch(IOException e) {}
  			return st.nextToken();
  		}
  		
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		long[] readArrayL(int n) {
  			long a[] = new long[n];
  			for(int i=0;i<n;i++) a[i] = nextLong();
  			return a;
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		
  		double[] readArrayD(int n) {
  			double a[] = new double[n];
  			for(int i=0;i<n;i++) a[i] = nextDouble();
  			return a;
  		}
  	}
  
  	static final Random random = new Random();
  
  	static void ruffleSort(int[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n), temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  	
  	static void ruffleSortL(long[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n);
  			long temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  }
  