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: #include <bits/stdc++.h>
  using namespace std;
  mt19937 ran(time(0));
  long long mod = 1e9 + 7;
  long long gcd(long long a, long long b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  long long Bpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) {
        ans *= a;
        ans %= mod;
      }
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long ans = 0;
    for (long long i = 0; i < n; i += 2) {
      long long now = 0;
      long long mib = 0;
      for (long long j = i + 1; j < n; j++) {
        long long b = now + a[i] - a[j];
        now += (j % 2 ? -a[j] : a[j]);
        if (b > 0 && b > a[i] || b < 0 && -b > a[j] || j % 2 == 0) {
          mib = min(mib, now);
          continue;
        }
        long long l = a[i], r = a[j];
        if (b > 0)
          l -= b;
        else
          r += b;
        if (l > 0 && r > 0 && l >= -mib) ans += min(l + (mib ? mib + 1 : 0), r);
        mib = min(mib, now);
      }
    }
    cout << ans;
  }
  