#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int P = 1000000007;
const int V = 1010;
int a[V];
long long s[V][V], L[V][V];
int n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) {
      if (i % 2 == 1)
        s[i][i] = a[i], L[i][i] = 0;
      else
        s[i][i] = -a[i], L[i][i] = -a[i];
      for (int j = i + 1; j <= n; ++j) {
        if (j % 2 == 1)
          s[i][j] = s[i][j - 1] + a[j];
        else
          s[i][j] = s[i][j - 1] - a[j];
        L[i][j] = min(L[i][j - 1], s[i][j]);
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; i += 2) {
      for (int j = i + 1; j <= n; j += 2) {
        if (j == i + 1) {
          ans += min(a[i], a[j]);
          continue;
        }
        long long M = L[i + 1][j - 1];
        long long wi = a[i];
        long long wj = a[j];
        if (s[i + 1][j - 1] < 0)
          wi += s[i + 1][j - 1], M -= s[i + 1][j - 1];
        else if (s[i + 1][j - 1] > 0)
          wj -= s[i + 1][j - 1];
        wi = min(wi, wi + M + 1);
        wj = min(wj, wj + M + 1);
        long long D = min(wi, wj);
        if (D < 0) continue;
        ans += D;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

<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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int P = 1000000007;
const int V = 1010;
int a[V];
long long s[V][V], L[V][V];
int n;
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) {
      if (i % 2 == 1)
        s[i][i] = a[i], L[i][i] = 0;
      else
        s[i][i] = -a[i], L[i][i] = -a[i];
      for (int j = i + 1; j <= n; ++j) {
        if (j % 2 == 1)
          s[i][j] = s[i][j - 1] + a[j];
        else
          s[i][j] = s[i][j - 1] - a[j];
        L[i][j] = min(L[i][j - 1], s[i][j]);
      }
    }
    long long ans = 0;
    for (int i = 1; i <= n; i += 2) {
      for (int j = i + 1; j <= n; j += 2) {
        if (j == i + 1) {
          ans += min(a[i], a[j]);
          continue;
        }
        long long M = L[i + 1][j - 1];
        long long wi = a[i];
        long long wj = a[j];
        if (s[i + 1][j - 1] < 0)
          wi += s[i + 1][j - 1], M -= s[i + 1][j - 1];
        else if (s[i + 1][j - 1] > 0)
          wj -= s[i + 1][j - 1];
        wi = min(wi, wi + M + 1);
        wj = min(wj, wj + M + 1);
        long long D = min(wi, wj);
        if (D < 0) continue;
        ans += D;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```