You are given an integer nn (n≥0n≥0) represented with kk digits in base (radix) bb. So,
n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.
For example, if b=17,k=3b=17,k=3 and a=[11,15,7]a=[11,15,7] then n=11⋅172+15⋅17+7=3179+255+7=3441n=11⋅172+15⋅17+7=3179+255+7=3441.
Determine whether nn is even or odd.
Input
The first line contains two integers bb and kk (2≤b≤1002≤b≤100, 1≤k≤1051≤k≤105) — the base of the number and the number of digits.
The second line contains kk integers a1,a2,…,aka1,a2,…,ak (0≤ai<b0≤ai<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=3⋅132+2⋅13+7=540n=3⋅132+2⋅13+7=540, which is even.
In the second example, n=123456789n=123456789 is odd.
In the third example, n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155 is odd.
In the fourth example n=2n=2.
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | #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; } |