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EVEN-ODD-QUERY Solution

/*
*  Author: Arpit Bhayani
*  https://arpitbhayani.me
*/
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <deque>
#include <iostream>
#include <list>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>

#define ll long long

#define MIN(a, b) a < b ? a : b
#define MAX(a, b) a > b ? a : b

using namespace std;

int i = 0;
char ch;
while((ch = getchar()) != '\n') {
str[i++] = ch;
}
str[i] = '\0';
return i;
}

int arr;

int npow(int a, int b) {
if(b == 0) {
//printf(" pow = %d - B\n", 1);
return 1;
}
//printf(" pow = %d - C\n", (a & 1) ? 1 : 2);
return (a & 1) ? 1 : 2;
}

int main(int argc, char *argv[]) {
int n, q;
int x, y;
char result = {"Even", "Odd"};

scanf("%d", &n);
for(int i = 1; i <= n ; i++ ) {
scanf("%d", &arr[i]);
}
scanf("%d", &q);

while(q--) {
int ans;

scanf("%d%d", &x, &y);

if(x < n && arr[x + 1] == 0 && x != y ) {
// Since the value just after "x"th index is 0
// so no matter what ans will be 1 -> odd
ans = 1;
}
else {
// In all other cases the odd/even of base determines
// the odd/even of result
ans = arr[x];
}

printf("%s\n", result[ans & 1]);
}

return 0;
}