Parity

You are given an integer nn (n0n0) represented with kk digits in base (radix) bb. So,

n=a1bk1+a2bk2+ak1b+ak.n=a1bk1+a2bk2+ak1b+ak.

For example, if b=17,k=3b=17,k=3 and a=[11,15,7]a=[11,15,7] then n=11172+1517+7=3179+255+7=3441n=11172+1517+7=3179+255+7=3441.

Determine whether nn is even or odd.

Input

The first line contains two integers bb and kk (2b1002b100, 1k1051k105) — the base of the number and the number of digits.

The second line contains kk integers a1,a2,,aka1,a2,,ak (0ai<b0ai<b) — the digits of nn.

The representation of nn contains no unnecessary leading zero. That is, a1a1 can be equal to 00 only if k=1k=1.

Output

Print “even” if nn 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=3132+213+7=540n=3132+213+7=540, which is even.

In the second example, n=123456789n=123456789 is odd.

In the third example, n=32994+92993+85992+7499+4=3164015155n=32994+92993+85992+7499+4=3164015155 is odd.

In the fourth example n=2n=2.

Solution:

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#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int b, k;
cin >> b >> k;
vector<int> a(k);
for (int i = 0; i < k; i++) {
    cin >> a[i];
}
int p = 1, s = 0;
for (int i = k - 1; i >= 0; i--) {
s = (s + p * a[i]) % 2;
p = p * b % 2;
}
cout << (s ? "odd" : "even") << '\n';
  return 0;
}