#include <bits/stdc++.h>
using namespace std;
clock_t TimeBegin, TimeEnd;
long long const INF = 1e18;
inline long long mod(string num, long long Base);
inline long long diMod(long long A, long long B, long long Base);
inline long long fpow(long long a, long long x, long long Base);
inline void Tstart();
inline void Tstop();
long long n = 0, b, k, a;
void Solve() {
  cin >> b >> k;
  for (long long i = (k - 1); i >= (0); --i) {
    cin >> a;
    n = (n + ((a % 2) * fpow(b, i, 2)) % 2) % 2;
  }
  cout << (n & 1 ? "odd" : "even");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Solve();
  return 0;
}
inline long long mod(string num, long long Base) {
  long long res = 0;
  for (long long i = 0; i < (num.length()); ++i)
    res = (res * 10 + (long long)num[i] - '0') % Base;
  return res;
}
inline long long fpow(long long a, long long x, long long Base) {
  if (x == 0) return 1;
  if (x & 1) {
    return a * fpow(a, x - 1, Base) % Base;
  } else {
    long long t = fpow(a, x / 2, Base);
    return t * t % Base;
  }
}
inline long long diMod(long long A, long long B, long long Base) {
  long long result = (A * fpow(B, Base - 2, Base)) % Base;
  return result;
}
inline void Tstart() { TimeBegin = clock(); }
inline void Tstop() {
  TimeEnd = clock();
  cout << "\nTime elapsed: " << 1000 * (TimeEnd - TimeBegin) / CLOCKS_PER_SEC
       << " ms\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;
clock_t TimeBegin, TimeEnd;
long long const INF = 1e18;
inline long long mod(string num, long long Base);
inline long long diMod(long long A, long long B, long long Base);
inline long long fpow(long long a, long long x, long long Base);
inline void Tstart();
inline void Tstop();
long long n = 0, b, k, a;
void Solve() {
  cin >> b >> k;
  for (long long i = (k - 1); i >= (0); --i) {
    cin >> a;
    n = (n + ((a % 2) * fpow(b, i, 2)) % 2) % 2;
  }
  cout << (n & 1 ? "odd" : "even");
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Solve();
  return 0;
}
inline long long mod(string num, long long Base) {
  long long res = 0;
  for (long long i = 0; i < (num.length()); ++i)
    res = (res * 10 + (long long)num[i] - '0') % Base;
  return res;
}
inline long long fpow(long long a, long long x, long long Base) {
  if (x == 0) return 1;
  if (x & 1) {
    return a * fpow(a, x - 1, Base) % Base;
  } else {
    long long t = fpow(a, x / 2, Base);
    return t * t % Base;
  }
}
inline long long diMod(long long A, long long B, long long Base) {
  long long result = (A * fpow(B, Base - 2, Base)) % Base;
  return result;
}
inline void Tstart() { TimeBegin = clock(); }
inline void Tstop() {
  TimeEnd = clock();
  cout << "\nTime elapsed: " << 1000 * (TimeEnd - TimeBegin) / CLOCKS_PER_SEC
       << " ms\n";
}

```