#include <bits/stdc++.h>
using namespace std;
long long x, y, z;
void memset_ll(long long a[], long long v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
void memset_s(char a[], char v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
int compare(const void* a, const void* b) {
  return (*(long long*)a - *(long long*)b);
}
void q_sort(long long a[], long long n) {
  qsort(a, n, sizeof(long long), compare);
}
static int compare1(const void* a, const void* b) {
  if (*(double*)a > *(double*)b)
    return 1;
  else if (*(double*)a < *(double*)b)
    return -1;
  else
    return 0;
}
void q_sort_d(double a[], long long n) {
  qsort(a, n, sizeof(double), compare1);
}
long long sum(long long n) {
  long long i = 0, a[6], j, k = 0;
  while (n > 0) {
    a[i] = n % 10;
    n /= 10;
    i++;
  }
  for (j = 0; j < i; j++) k += a[j];
  return k;
}
string trans_upper(string ss) {
  transform(ss.begin(), ss.end(), ss.begin(), ::toupper);
  return ss;
}
long long power(long long a, long long b) {
  long long pow = 1;
  while (b) {
    if (b & 1) {
      pow = pow * a;
      --b;
    }
    a = a * a;
    b = b / 2;
  }
  return pow;
}
long long fact(long long n) {
  long long f = 1, i;
  for (i = 1; i <= n; i++) f = f * i;
  return f;
}
int main() {
  long long t, l, p, q, r, i, j, h, k, e = 0, f = 0, g = 0;
  long long temp = 0, n, m, d = 0, nl, np;
  cin >> n >> m;
  long long a[m];
  for (i = 0; i < m; i++) {
    cin >> a[i];
    j = i + 1;
    k = m - j;
    q = power(n, k);
    p = a[i] * q;
    d += p;
  }
  if (d % 2 == 0)
    cout << "even";
  else
    cout << "odd";
  return 0;
}

### Prompt

Your task is to create 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;
long long x, y, z;
void memset_ll(long long a[], long long v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
void memset_s(char a[], char v, long long n) {
  for (long long i = 0; i < n; i++) a[i] = v;
}
int compare(const void* a, const void* b) {
  return (*(long long*)a - *(long long*)b);
}
void q_sort(long long a[], long long n) {
  qsort(a, n, sizeof(long long), compare);
}
static int compare1(const void* a, const void* b) {
  if (*(double*)a > *(double*)b)
    return 1;
  else if (*(double*)a < *(double*)b)
    return -1;
  else
    return 0;
}
void q_sort_d(double a[], long long n) {
  qsort(a, n, sizeof(double), compare1);
}
long long sum(long long n) {
  long long i = 0, a[6], j, k = 0;
  while (n > 0) {
    a[i] = n % 10;
    n /= 10;
    i++;
  }
  for (j = 0; j < i; j++) k += a[j];
  return k;
}
string trans_upper(string ss) {
  transform(ss.begin(), ss.end(), ss.begin(), ::toupper);
  return ss;
}
long long power(long long a, long long b) {
  long long pow = 1;
  while (b) {
    if (b & 1) {
      pow = pow * a;
      --b;
    }
    a = a * a;
    b = b / 2;
  }
  return pow;
}
long long fact(long long n) {
  long long f = 1, i;
  for (i = 1; i <= n; i++) f = f * i;
  return f;
}
int main() {
  long long t, l, p, q, r, i, j, h, k, e = 0, f = 0, g = 0;
  long long temp = 0, n, m, d = 0, nl, np;
  cin >> n >> m;
  long long a[m];
  for (i = 0; i < m; i++) {
    cin >> a[i];
    j = i + 1;
    k = m - j;
    q = power(n, k);
    p = a[i] * q;
    d += p;
  }
  if (d % 2 == 0)
    cout << "even";
  else
    cout << "odd";
  return 0;
}

```