#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;
}

### Prompt

Generate a Cpp solution to the following problem:

<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.

### Response

```cpp
#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;
}

```