#include <bits/stdc++.h>
using namespace std;
bool sortBySecond(const pair<long long, long long> &a,
                  const pair<long long, long long> b) {
  return a.second > b.second;
}
int ncr[1001][1001];
bool vis_ncr[1001][1001];
long long NCR(int n, int r) {
  if (n < r) return 0;
  if (vis_ncr[n][r]) return ncr[n][r];
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  ncr[n][r] = NCR(n - 1, r) + NCR(n - 1, r - 1);
  vis_ncr[n][r] = true;
  return ncr[n][r];
}
bool isPrime[100005];
void _seive() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  isPrime[0] = false;
  for (int i = 2; i * i <= 100005; i++) {
    for (int j = i * i; j <= 100005; j += i) {
      isPrime[j] = false;
    }
  }
}
long long expo(int a, int b) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else if (b % 2 == 0) {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    return ret;
  } else {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    ret = (ret % 1000000007 * a % 1000000007) % 1000000007;
    return ret;
  }
}
long long inverseMod(long long a, long long m) { return expo(a, m - 2) % m; }
long long fact[100005];
void factorial() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i < 100005; i++)
    fact[i] = (fact[i - 1] % 1000000007 * i % 1000000007) % 1000000007;
}
long long GCD(long long a, long long b) {
  if (a == 0) return b;
  return GCD(b % a, a);
}
long long mult(long long a, long long b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) return a - 1000000007;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + 1000000007;
  return a;
}
void _run() {
  int b, k;
  cin >> b >> k;
  int a[k];
  for (int i = 0; i < k; i++) cin >> a[i];
  if (b & 1) {
    int cnt = 0;
    for (int i = 0; i < k; i++)
      if (a[i] & 1) cnt++;
    if (cnt % 2 == 0)
      cout << "even" << endl;
    else
      cout << "odd" << endl;
    return;
  }
  cout << (a[k - 1] & 1 ? "odd" : "even") << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int test = 1;
  while (test--) _run();
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
bool sortBySecond(const pair<long long, long long> &a,
                  const pair<long long, long long> b) {
  return a.second > b.second;
}
int ncr[1001][1001];
bool vis_ncr[1001][1001];
long long NCR(int n, int r) {
  if (n < r) return 0;
  if (vis_ncr[n][r]) return ncr[n][r];
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  ncr[n][r] = NCR(n - 1, r) + NCR(n - 1, r - 1);
  vis_ncr[n][r] = true;
  return ncr[n][r];
}
bool isPrime[100005];
void _seive() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  isPrime[0] = false;
  for (int i = 2; i * i <= 100005; i++) {
    for (int j = i * i; j <= 100005; j += i) {
      isPrime[j] = false;
    }
  }
}
long long expo(int a, int b) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else if (b % 2 == 0) {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    return ret;
  } else {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000007 * ret % 1000000007) % 1000000007;
    ret = (ret % 1000000007 * a % 1000000007) % 1000000007;
    return ret;
  }
}
long long inverseMod(long long a, long long m) { return expo(a, m - 2) % m; }
long long fact[100005];
void factorial() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i < 100005; i++)
    fact[i] = (fact[i - 1] % 1000000007 * i % 1000000007) % 1000000007;
}
long long GCD(long long a, long long b) {
  if (a == 0) return b;
  return GCD(b % a, a);
}
long long mult(long long a, long long b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) return a - 1000000007;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + 1000000007;
  return a;
}
void _run() {
  int b, k;
  cin >> b >> k;
  int a[k];
  for (int i = 0; i < k; i++) cin >> a[i];
  if (b & 1) {
    int cnt = 0;
    for (int i = 0; i < k; i++)
      if (a[i] & 1) cnt++;
    if (cnt % 2 == 0)
      cout << "even" << endl;
    else
      cout << "odd" << endl;
    return;
  }
  cout << (a[k - 1] & 1 ? "odd" : "even") << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int test = 1;
  while (test--) _run();
  return 0;
}

```