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 java.io.*;
  import java.math.*;
  import java.awt.geom.*;
  
  import static java.lang.Math.*;
  
  public class Solution implements Runnable {
  
  	long mod1 = (long) 1e9 + 7;
  	int mod2 = 998244353;
  
  	public void solve() throws Exception {
  		int n=sc.nextInt();
  		long arr[] = new long[n];
  		for(int i=0;i<n;i++) {
  			arr[i]=sc.nextLong();
  		}
  		long ans=0;
  		for(int i=0;i<n;i++) {
  			if(i%2==1) {
  				long currClosing = arr[i];
  				long initial = arr[i];
  				long min=arr[i];
  				for(int j=i-1;j>=0;j--) {
  					if(min<0) break;
  					if((i-j)%2==0) {
  						currClosing += arr[j];
  					} else {
  						if(arr[j]<=currClosing) {
  							if(min==arr[i]) {
  								ans += arr[j];
  								currClosing -= arr[j];
  								min = min-arr[j];
  								continue;
  							}
  							long leftWith = currClosing - min;
  							long nn = arr[j];
  							nn -= leftWith;
  							if(nn>=0)
  							ans += nn+1;
  							currClosing -= arr[j];
  							min=Math.min(min, currClosing);
  						}
  						else {
  							if(j==i-1)
  							ans += min;
  							else ans += min+1;
  							break;
  						}
  					}
  					
  				}
  			}
  		}
  		out.println(ans);
  		
  
  	}
  
  	static long gcd(long a, long b) {
  		if (a == 0)
  			return b;
  		return gcd(b % a, a);
  	}
  
  	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 long ncr(int n, int r, long p) {
  		if (r > n)
  			return 0l;
  		if (r > n - r)
  			r = n - r;
  
  		long C[] = new long[r + 1];
  
  		C[0] = 1;
  
  		for (int i = 1; i <= n; i++) {
  
  			for (int j = Math.min(i, r); j > 0; j--)
  				C[j] = (C[j] + C[j - 1]) % p;
  		}
  		return C[r] % p;
  	}
  
  	void sieveOfEratosthenes(boolean prime[], int size) {
  		for (int i = 0; i < size; i++)
  			prime[i] = true;
  
  		for (int p = 2; p * p < size; p++) {
  			if (prime[p] == true) {
  				for (int i = p * p; i < size; i += p)
  					prime[i] = false;
  			}
  		}
  	}
  
  	static int LowerBound(int a[], int x) { // smallest index having value >= x
  		int l = -1, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] >= x)
  				r = m;
  			else
  				l = m;
  		}
  		return r;
  	}
  
  	static int UpperBound(int a[], int x) {// biggest index having value <= x
  		int l = -1, r = a.length;
  		while (l + 1 < r) {
  			int m = (l + r) >>> 1;
  			if (a[m] <= x)
  				l = m;
  			else
  				r = m;
  		}
  		return l + 1;
  	}
  
  	public long power(long x, long y, long p) {
  		long res = 1;
  		// out.println(x+" "+y);
  		x = x % p;
  		if (x == 0)
  			return 0;
  
  		while (y > 0) {
  			if ((y & 1) == 1)
  				res = (res * x) % p;
  			y = y >> 1;
  			x = (x * x) % p;
  		}
  		return res;
  	}
  
  	static Throwable uncaught;
  
  	BufferedReader in;
  	FastScanner sc;
  	PrintWriter out;
  
  	@Override
  	public void run() {
  		try {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(System.out);
  			sc = new FastScanner(in);
  			solve();
  		} catch (Throwable uncaught) {
  			Solution.uncaught = uncaught;
  		} finally {
  			out.close();
  		}
  	}
  
  	public static void main(String[] args) throws Throwable {
  		Thread thread = new Thread(null, new Solution(), "", (1 << 26));
  		thread.start();
  		thread.join();
  		if (Solution.uncaught != null) {
  			throw Solution.uncaught;
  		}
  	}
  
  }
  
  class FastScanner {
  
  	BufferedReader in;
  	StringTokenizer st;
  
  	public FastScanner(BufferedReader in) {
  		this.in = in;
  	}
  
  	public String nextToken() throws Exception {
  		while (st == null || !st.hasMoreTokens()) {
  			st = new StringTokenizer(in.readLine());
  		}
  		return st.nextToken();
  	}
  
  	public int nextInt() throws Exception {
  		return Integer.parseInt(nextToken());
  	}
  
  	public int[] readArray(int n) throws Exception {
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] = nextInt();
  		return a;
  	}
  
  	public long nextLong() throws Exception {
  		return Long.parseLong(nextToken());
  	}
  
  	public double nextDouble() throws Exception {
  		return Double.parseDouble(nextToken());
  	}
  
  }