#include <bits/stdc++.h>
using namespace std;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[100005], a(2000005), f(200005, 0), d(200005, 0),
    b(1000005);
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int b, k;
  cin >> b >> k;
  long long int c = (b % 2), c1 = 0;
  vector<long long int> a(k);
  for (long long int i = 0; i < k; i++) {
    cin >> a[i];
    if (i == k - 1) {
      if (a[i] % 2) c1++;
    } else {
      if (c == 1 && a[i] % 2) c1++;
    }
  }
  if (c1 % 2) {
    cout << "odd"
         << "\n";
  } else {
    cout << "even"
         << "\n";
  }
}

### Prompt

Please provide a cpp coded solution 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;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[100005], a(2000005), f(200005, 0), d(200005, 0),
    b(1000005);
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int b, k;
  cin >> b >> k;
  long long int c = (b % 2), c1 = 0;
  vector<long long int> a(k);
  for (long long int i = 0; i < k; i++) {
    cin >> a[i];
    if (i == k - 1) {
      if (a[i] % 2) c1++;
    } else {
      if (c == 1 && a[i] % 2) c1++;
    }
  }
  if (c1 % 2) {
    cout << "odd"
         << "\n";
  } else {
    cout << "even"
         << "\n";
  }
}

```