For an array b of length m we define the function f as 

 f(b) = \begin{cases} b[1] &   if  m = 1 \\\ f(b[1] ⊕ b[2],b[2] ⊕ b[3],...,b[m-1] ⊕ b[m]) &   otherwise, \end{cases}  

where ⊕ is [bitwise exclusive OR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

For example, f(1,2,4,8)=f(1⊕2,2⊕4,4⊕8)=f(3,6,12)=f(3⊕6,6⊕12)=f(5,10)=f(5⊕10)=f(15)=15

You are given an array a and a few queries. Each query is represented as two integers l and r. The answer is the maximum value of f on all continuous subsegments of the array a_l, a_{l+1}, …, a_r.

Input

The first line contains a single integer n (1 ≤ n ≤ 5000) — the length of a.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30}-1) — the elements of the array.

The third line contains a single integer q (1 ≤ q ≤ 100 000) — the number of queries.

Each of the next q lines contains a query represented as two integers l, r (1 ≤ l ≤ r ≤ n).

Output

Print q lines — the answers for the queries.

Examples

Input

3
8 4 1
2
2 3
1 2


Output

5
12


Input

6
1 2 4 8 16 32
4
1 6
2 5
3 4
1 2


Output

60
30
12
3

Note

In first sample in both queries the maximum value of the function is reached on the subsegment that is equal to the whole segment.

In second sample, optimal segment for first query are [3,6], for second query — [2,5], for third — [3,4], for fourth — [1,2].
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int n;
int a[5005];
int dp[5005][5005];
int dp2[5005][5005];
int recur(int posl, int posr) {
  if (posl == posr) return a[posl];
  if (dp[posl][posr] != -1) return dp[posl][posr];
  int ret1, ret2, ret;
  ret1 = recur(posl, posr - 1);
  ret2 = recur(posl + 1, posr);
  ret = (ret1 ^ ret2);
  dp[posl][posr] = ret;
  return ret;
}
int recur2(int posl, int posr) {
  if (posl == posr) return a[posl];
  if (dp2[posl][posr] != -1) return dp2[posl][posr];
  int ret1, ret2, ret3, ret;
  ret1 = recur2(posl, posr - 1);
  ret2 = recur2(posl + 1, posr);
  ret3 = recur(posl, posr);
  ret = max(ret1, max(ret2, ret3));
  dp2[posl][posr] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  memset(dp, -1, sizeof(dp));
  memset(dp2, -1, sizeof(dp2));
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  int q;
  cin >> q;
  for (int i = 0, l, r; i < q; ++i) {
    cin >> l >> r;
    --l, --r;
    cout << recur2(l, r) << '\n';
  }
}
