#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
const int MOD = 1e9 + 7;
template <class T>
class my_array {
 private:
  int n;
  vector<T> A;

 public:
  my_array(int k) {
    n = k;
    A.resize(n);
  }
  ~my_array(void){};
  void get(void) {
    for (int i = 0; i < n; i++) cin >> A.at(i);
  }
  void sorted(bool flag = true) {
    if (flag)
      sort(A.begin(), A.end());
    else
      sort(A.begin(), A.end(), greater<T>());
  }
  void show(void) {
    for (int i = 0; i < n - 1; i++) cout << A.at(i) << " ";
    cout << A.at(n - 1) << "\n";
  }
  int solve(int a, int b) {
    if (a >= b) return 0;
    int mn_idx = 0, min = 1e9;
    for (int i = a; i < b; i++)
      if (min > A.at(i)) {
        min = A.at(i);
        mn_idx = i;
      }
    for (int i = a; i < b; i++) A.at(i) -= min;
    return solve(a, mn_idx) + min + solve(mn_idx + 1, b);
  }
};
int main(void) {
  int a, b, k, res = 0;
  cin >> b >> k;
  for (int i = 0; i < k - 1; i++) {
    cin >> a;
    if ((a & 1) && (b & 1)) res++;
  }
  cin >> a;
  if ((a & 1)) res++;
  cout << ((res & 1) ? "odd" : "even") << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,

$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$

For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.

Determine whether n is even or odd.

Input

The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.

The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.

The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.

Output

Print "even" if n is even, otherwise print "odd".

You can print each letter in any case (upper or lower).

Examples

Input

13 3
3 2 7


Output

even


Input

10 9
1 2 3 4 5 6 7 8 9


Output

odd


Input

99 5
32 92 85 74 4


Output

odd


Input

2 2
1 0


Output

even

Note

In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.

In the second example, n = 123456789 is odd.

In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.

In the fourth example n = 2.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
const int MOD = 1e9 + 7;
template <class T>
class my_array {
 private:
  int n;
  vector<T> A;

 public:
  my_array(int k) {
    n = k;
    A.resize(n);
  }
  ~my_array(void){};
  void get(void) {
    for (int i = 0; i < n; i++) cin >> A.at(i);
  }
  void sorted(bool flag = true) {
    if (flag)
      sort(A.begin(), A.end());
    else
      sort(A.begin(), A.end(), greater<T>());
  }
  void show(void) {
    for (int i = 0; i < n - 1; i++) cout << A.at(i) << " ";
    cout << A.at(n - 1) << "\n";
  }
  int solve(int a, int b) {
    if (a >= b) return 0;
    int mn_idx = 0, min = 1e9;
    for (int i = a; i < b; i++)
      if (min > A.at(i)) {
        min = A.at(i);
        mn_idx = i;
      }
    for (int i = a; i < b; i++) A.at(i) -= min;
    return solve(a, mn_idx) + min + solve(mn_idx + 1, b);
  }
};
int main(void) {
  int a, b, k, res = 0;
  cin >> b >> k;
  for (int i = 0; i < k - 1; i++) {
    cin >> a;
    if ((a & 1) && (b & 1)) res++;
  }
  cin >> a;
  if ((a & 1)) res++;
  cout << ((res & 1) ? "odd" : "even") << "\n";
  return 0;
}

```