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 inf = 0x3f3f3f3f;
  const long long INF = 1e18;
  const int maxn = 2e5 + 7;
  const long long mod = 1e9 + 7;
  inline long long read() {
    long long x = 0;
    bool f = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
    while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  void out(long long x) {
    int stackk[20];
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (!x) {
      putchar('0');
      return;
    }
    int top = 0;
    while (x) stackk[++top] = x % 10, x /= 10;
    while (top) putchar(stackk[top--] + '0');
  }
  long long qpow(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  long long n, a[maxn];
  int main() {
    n = read();
    for (int i = (1); i <= (n); ++i) a[i] = read();
    long long ans = 0;
    for (int i = 1; i <= n; i += 2) {
      long long zuo = a[i];
      long long duo = 0;
      for (int j = i + 1; j <= n; j++) {
        if (j % 2)
          duo += a[j];
        else {
          long long tmp = a[j];
          long long mn = min(tmp, duo);
          tmp -= mn, duo -= mn;
          if (duo == 0) {
            mn = min(tmp, zuo);
            zuo -= mn, tmp -= mn;
            ans += mn;
            if (j != i + 1) ans++;
            if (tmp) break;
          }
        }
      }
    }
    out(ans);
    return 0;
  }
  