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;
  const int N = 400010;
  const int mod = 1e9 + 7;
  const int INF = 1e9 + 7;
  const double Pi = acos(-1.0);
  int t, n;
  long long a[N];
  int dir[][2] = {{0, 0},  {0, 1},   {-1, 0}, {0, -1}, {1, 0},
                  {-1, 1}, {-1, -1}, {1, -1}, {1, 1}};
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  inline int gcd(int x, int y) {
    if (!x) return y;
    return gcd(y % x, x);
  }
  inline int power(int x, int k, int p) {
    int res = 1;
    for (; k; k >>= 1, x = (long long)x * x % p)
      if (k & 1) res = (long long)res * x % p;
    return res;
  }
  long long k[N];
  void solve() {
    n = read();
    long long r = 0, ans = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      if (i % 2) {
        r += a[i];
      } else {
        ans += min(r, a[i]);
        r = max((long long)0, r - a[i]);
        k[i] = 1;
        for (int j = i - 1; j >= 1; j -= 2) {
          if (a[j]) {
            int tmp = a[j];
            a[j] = max((long long)0, a[j] - a[i]);
            a[i] = max((long long)0, a[i] - tmp);
          }
          int las = j;
          if (a[i] == 0) {
            while (j >= 1 && a[j] == 0) j -= 2;
            for (int l = j; l < i; ++l) {
              ans += k[l];
            }
          }
          if (a[i] == 0) {
            for (int l = las; l < i; ++l) {
              k[l] = 0;
            }
            break;
          }
        }
        if (a[i] != 0) {
          for (int l = 1; l < i; ++l) {
            ans += k[l];
            k[l] = 0;
          }
          k[i] = 0;
        }
      }
    }
    cout << ans << endl;
    return;
  }
  int main() {
    solve();
    return 0;
  }
  