#include <bits/stdc++.h>
using namespace std;
double eps = 0.0000001;
double pi = acos(-1);
long long fastpower(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = (ans * b);
    }
    b = b * b;
    p /= 2;
  }
  return ans;
}
long long fastpowermod(long long b, long long p, long long m) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = ((ans % m) * (b % m)) % m;
    }
    b = ((b % m) * (b % m)) % m;
    p /= 2;
  }
  return ans % m;
}
string makeitstring(long long n) {
  string ans;
  bool ch = 0;
  if (n < 0) {
    n *= -1;
    ch = 1;
  }
  if (n == 0) {
    ans = "0";
  };
  while (n) {
    long long mo = n % 10;
    mo += 48;
    char m = mo;
    ans = m + ans;
    n /= 10;
  }
  if (ch) {
    ans = '-' + ans;
  }
  return ans;
}
bool compare(double f, double s) { return (abs(f - s) < eps); }
long long gcd(long long a, long long b) {
  while (b != 0) {
    long long a2 = a;
    a = b;
    b = a2 % b;
  }
  return a;
}
const int N = 1e2 + 1;
vector<bool> v;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long b, k;
  cin >> b >> k;
  int even = 0;
  int odd = 0;
  for (int i = 0; i < k; i++) {
    long long x;
    cin >> x;
    if (i == k - 1) {
      v.push_back(x % 2);
      break;
    }
    if (b % 2 && x % 2) {
      v.push_back(1);
    } else {
      v.push_back(0);
    }
  }
  bool ans = v[0];
  for (int i = 1; i < v.size(); i++) {
    if (ans == 0 && v[i] == 1) {
      ans = 1;
    } else if (ans == 1 && v[i] == 1) {
      ans = 0;
    }
  }
  if (ans) {
    cout << "odd";
  } else {
    cout << "even";
  }
  return 0;
}

### Prompt

Generate 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;
double eps = 0.0000001;
double pi = acos(-1);
long long fastpower(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = (ans * b);
    }
    b = b * b;
    p /= 2;
  }
  return ans;
}
long long fastpowermod(long long b, long long p, long long m) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = ((ans % m) * (b % m)) % m;
    }
    b = ((b % m) * (b % m)) % m;
    p /= 2;
  }
  return ans % m;
}
string makeitstring(long long n) {
  string ans;
  bool ch = 0;
  if (n < 0) {
    n *= -1;
    ch = 1;
  }
  if (n == 0) {
    ans = "0";
  };
  while (n) {
    long long mo = n % 10;
    mo += 48;
    char m = mo;
    ans = m + ans;
    n /= 10;
  }
  if (ch) {
    ans = '-' + ans;
  }
  return ans;
}
bool compare(double f, double s) { return (abs(f - s) < eps); }
long long gcd(long long a, long long b) {
  while (b != 0) {
    long long a2 = a;
    a = b;
    b = a2 % b;
  }
  return a;
}
const int N = 1e2 + 1;
vector<bool> v;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long b, k;
  cin >> b >> k;
  int even = 0;
  int odd = 0;
  for (int i = 0; i < k; i++) {
    long long x;
    cin >> x;
    if (i == k - 1) {
      v.push_back(x % 2);
      break;
    }
    if (b % 2 && x % 2) {
      v.push_back(1);
    } else {
      v.push_back(0);
    }
  }
  bool ans = v[0];
  for (int i = 1; i < v.size(); i++) {
    if (ans == 0 && v[i] == 1) {
      ans = 1;
    } else if (ans == 1 && v[i] == 1) {
      ans = 0;
    }
  }
  if (ans) {
    cout << "odd";
  } else {
    cout << "even";
  }
  return 0;
}

```