#include <bits/stdc++.h>
using namespace std;
static const double EPS = 1e-9;
template <class T>
bool INRANGE(T x, T a, T b) {
  return a <= x && x <= b;
}
template <class T>
void amin(T &a, T v) {
  if (a > v) a = v;
}
template <class T>
void amax(T &a, T v) {
  if (a < v) a = v;
}
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) {
  return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y)));
}
double SQSUM(double x, double y) { return x * x + y * y; }
int main() {
  int b, k;
  scanf("%d %d ", &b, &k);
  vector<int> a(k);
  for (int i = 0; i < k; ++i) {
    scanf("%d ", &a[i]);
  }
  int even = 0;
  int odd = 0;
  for (int i = 0; i < k; ++i) {
    if (i == k - 1) {
      if (a[i] % 2 == 1) {
        odd++;
      } else {
        even++;
      }
    } else {
      if (a[i] % 2 == 1 && b % 2 == 1) {
        odd++;
      } else {
        even++;
      }
    }
  }
  if (odd % 2 == 1) {
    printf("odd\n");
  } else {
    printf("even\n");
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
static const double EPS = 1e-9;
template <class T>
bool INRANGE(T x, T a, T b) {
  return a <= x && x <= b;
}
template <class T>
void amin(T &a, T v) {
  if (a > v) a = v;
}
template <class T>
void amax(T &a, T v) {
  if (a < v) a = v;
}
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) {
  return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y)));
}
double SQSUM(double x, double y) { return x * x + y * y; }
int main() {
  int b, k;
  scanf("%d %d ", &b, &k);
  vector<int> a(k);
  for (int i = 0; i < k; ++i) {
    scanf("%d ", &a[i]);
  }
  int even = 0;
  int odd = 0;
  for (int i = 0; i < k; ++i) {
    if (i == k - 1) {
      if (a[i] % 2 == 1) {
        odd++;
      } else {
        even++;
      }
    } else {
      if (a[i] % 2 == 1 && b % 2 == 1) {
        odd++;
      } else {
        even++;
      }
    }
  }
  if (odd % 2 == 1) {
    printf("odd\n");
  } else {
    printf("even\n");
  }
  return 0;
}

```