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.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class C {
  
  	public static void main(String[] args) {
  		FastScanner fs=new FastScanner();
  //		int T=fs.nextInt();
  		int T=1;
  		for (int tt=0; tt<T; tt++) {
  			int n=fs.nextInt();
  			int[] a=fs.readArray(n);
  			boolean isOpen=true;
  			long prefixSum=0;
  			ArrayList<Range> ranges=new ArrayList<>();
  			long ans=0;
  			
  			for (int i:a) {
  				if (isOpen) {
  					Range r=new Range(prefixSum, prefixSum+i-1);
  					ranges.add(r);
  					prefixSum+=i;
  				}
  				else {
  					long biggest=prefixSum-1;
  					long smallest=prefixSum-i;
  					prefixSum-=i;
  					for (Range r:ranges) {
  						ans+=r.getOverlapAndTrim(smallest, biggest);
  					}
  				}
  				isOpen^=true;
  			}
  			System.out.println(ans);
  		}
  	}
  	
  	static class Range {
  		long start, end;
  		boolean dead=false;
  		public Range(long start, long end) {
  			this.start=start;
  			this.end=end;
  		}
  		public long getOverlapAndTrim(long l, long r) {
  			if (end<start) return 0;
  			
  			long rr=Math.min(end, r);
  			long ll=Math.max(l, start);
  			end=Math.min(end, l);
  			long ans=Math.max(rr-ll+1, 0);
  //			System.out.println("Returning "+ans+" for lr: "+l+" "+r+" on "+start+" "+end);
  			return ans;
  		}
  	}
  	
  	/*
  	range += 1
  	range sum
  	 */
  //	static class ST {
  //		long leftmost, rightmost;
  //		ST lChild, rChild;
  //		long toProp;
  //		long sum;
  //		
  //		public ST(long l, long r) {
  //			leftmost=l;
  //			rightmost=r;
  //		}
  //		void recalc() {
  //			if (leftmost!=rightmost)
  //				sum=lChild.sum()+rChild.sum();
  //		}
  //		void prop() {
  //			if (toProp==0 || leftmost==rightmost) return;
  //			makeKids();
  //			lChild.toProp+=toProp;
  //			rChild.toProp+=toProp;
  //			toProp=0;
  //			recalc();
  //		}
  //		long sum() {
  //			return sum+toProp*(rightmost-leftmost+1);
  //		}
  //		void rangeAdd(long l, long r, int d) {
  //			if (l>rightmost || r<leftmost) return;
  //			if (l<=leftmost && r>=rightmost) {
  //				toProp+=d;
  //			}
  //			makeKids();
  //			prop();
  //			lChild.rangeAdd(l, r, d);
  //			rChild.rangeAdd(l, r, d);
  //			recalc();
  //		}
  //		long rangeSum(long l, long r) {
  //			if (l>rightmost || r<leftmost) return 0;
  //			if (l<=leftmost && r>=rightmost) {
  //				return sum();
  //			}
  //			makeKids();
  //			prop();
  //			return lChild.rangeSum(l, r)+rChild.rangeSum(l, r);
  //		}
  //		
  //		void makeKids() {
  //			if (lChild!=null || (leftmost==rightmost)) return;
  //			long mid=(leftmost+rightmost)/2;
  //			lChild=new ST(leftmost, mid);
  //			rChild=new ST(mid+1, rightmost);
  //		}
  //	}
  
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		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());
  		}
  	}
  
  	
  }
  