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 Codeforces {
  	public static void main(String[] args) throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		//int cases = Integer.parseInt(br.readLine());
  		//o:while(cases-- > 0) {
  			String[] str = br.readLine().split(" ");
  			int n = Integer.parseInt(str[0]);
  			int[] c = new int[n];
  			str = br.readLine().split(" ");
  			for(int i=0; i<n; i++) {
  				c[i] = Integer.parseInt(str[i]);
  			}
  			long ans = 0;
  			for(int i=0; i+1<n; i+=2) { // ck - open,, ck1 or ck+1 - close
  				ans += Math.min(c[i], c[i+1]);
  				if(c[i+1] > c[i]) { // cant go further if there r pending close brackets here itself
  					continue;
  				}
  				long ci = c[i] - c[i+1]; // pending opens at i
  				long cj = 0; // represents the pending opens till the j or j+1 from i+2
  				for(int j=i+2; j+1<n; j+=2) {
  					cj += c[j];
  					if(cj > c[j+1]) { // pending opens > closes available here, so no points here
  						// close how many ever u can and move on
  						cj -= c[j+1];
  						continue;
  					}
  					// we can close some ci if ci arent 0.
  					long cj1 = c[j+1] - cj; // spare ones left after closing all opens not including i opens
  					cj = 0; // since we just closed all pending opens except at i
  					ans++; // we can consider i to j+1 as regular, so ans++ even if cj1 == 0
  					if(cj1 > ci) {
  						ans += ci;
  						break; // if we have closes even after we close opens at i, we cant move further
  					}
  					ci -= cj1; // might become 0 or stay positive
  					ans += cj1;
  				}
  			}
  			System.out.println(ans);
  		//}
  	}
  }