#include <bits/stdc++.h>
using namespace std;
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long minize(long long& a, long long b) {
  return a = (a > b ? b : a);
}
inline long long maxize(long long& a, long long b) {
  return a = (a < b ? b : a);
}
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e6;
long long A[MAXN + 5];
long long Solves() {
  long long n, m, k, cnt = 0, ans = 0, x, y, q, c, sum = 0, v, t;
  cin >> k >> n;
  for (long long i = 1; i <= n; i++) cin >> A[i];
  if (k % 2 == 0) {
    return cout << (A[n] & 1 ? "odd" : "even") << endl, 0;
  }
  for (long long i = 1; i <= n; i++) {
    sum += A[i] & 1;
  }
  cout << (sum & 1 ? "odd" : "even") << endl;
  return 0;
}
int main() {
  long long JUDGE_ONLINE = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  for (long long T = 1; T <= test; T++) {
    Solves();
  }
  if (!JUDGE_ONLINE)
    cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}

### Prompt

Create a solution in cpp for the following problem:

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;
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline long long minize(long long& a, long long b) {
  return a = (a > b ? b : a);
}
inline long long maxize(long long& a, long long b) {
  return a = (a < b ? b : a);
}
inline string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
double const eps = 1e-6;
long long const Base = 1e9 + 7, oo = 1e17, MAXN = 1e6;
long long A[MAXN + 5];
long long Solves() {
  long long n, m, k, cnt = 0, ans = 0, x, y, q, c, sum = 0, v, t;
  cin >> k >> n;
  for (long long i = 1; i <= n; i++) cin >> A[i];
  if (k % 2 == 0) {
    return cout << (A[n] & 1 ? "odd" : "even") << endl, 0;
  }
  for (long long i = 1; i <= n; i++) {
    sum += A[i] & 1;
  }
  cout << (sum & 1 ? "odd" : "even") << endl;
  return 0;
}
int main() {
  long long JUDGE_ONLINE = 1;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  for (long long T = 1; T <= test; T++) {
    Solves();
  }
  if (!JUDGE_ONLINE)
    cout << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
}

```