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 rnd(time(0));
  bool isPrime(long long a) {
    long long i;
    for (i = 2; i * i <= a; i++) {
      if (a % i == 0) {
        return 0;
      }
    }
    return (a != 1);
  }
  bool isPalindrom(const string& s, int l, int r) {
    for (int i = 0; i < (r - l + 1) / 2; i++) {
      if (s[l + i] != s[r - i]) {
        return 0;
      }
    }
    return 1;
  }
  template <class type1>
  type1 gcd(type1 a, type1 b) {
    type1 t;
    while (b) {
      t = b;
      b = a % b;
      a = t;
    }
    return a;
  }
  long long positiveMod(long long v, long long md) {
    v = v % md;
    if (v < 0) {
      v += md;
    }
    return v;
  }
  vector<long long> dat;
  vector<vector<long long> > infMin;
  vector<vector<long long> > infMax;
  vector<vector<long long> > infScore;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long n, i, j, a, b, c;
    unsigned long long res = 0;
    cin >> n;
    dat.push_back(0);
    for (i = 0; i < n; i++) {
      cin >> a;
      dat.push_back(a);
    }
    infMin.resize(n + 2, vector<long long>(n + 2, 0));
    infMax.resize(n + 2, vector<long long>(n + 2, 0));
    infScore.resize(n + 2, vector<long long>(n + 2, 0));
    for (i = 1; i <= n; i++) {
      long long val = 0;
      for (j = i; j <= n; j++) {
        if (j % 2 == 1) {
          val += dat[j];
        } else {
          val -= dat[j];
        }
        infMin[i][j] = min(infMin[i][j - 1], val);
        infMax[i][j] = max(infMax[i][j - 1], val);
        infScore[i][j] = val;
      }
    }
    for (i = 1; i <= n; i++) {
      for (j = i + 1; j <= n; j++) {
        long long mn = infMin[i + 1][j - 1], mx = infMin[i + 1][j - 1],
                  sc = infScore[i + 1][j - 1];
        long long sp = 0;
        if (i % 2 == 0) {
          continue;
        }
        if (j % 2 == 1) {
          continue;
        }
        long long v1 = dat[i];
        long long v2 = dat[j];
        v1 -= abs(mn);
        sc += abs(mn);
        sp += abs(mn);
        if (v1 < 0) {
          continue;
        }
        v2 -= sc;
        sp += sc;
        if (v2 < 0) {
          continue;
        }
        res += min(v1, v2);
        if (sp != 0) {
          res++;
        }
      }
    }
    cout << res << "\n";
    return 0;
  }
  