#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,fma,tune=native")
using namespace std;
const int MAXN = 5e6 + 10;
const int MAXM = 2e6 + 10;
const long long MOD = 1e9 + 7;
const long long INF = 0x3f3f3f3f;
const unsigned long long BASE = 233;
const int base = 5;
mt19937_64 mt(time(0));
char rubbish[110];
bool GETDATA;
inline long long read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
inline void write(long long x, char con = '\n') {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10, 0);
  putchar(x % 10 + '0');
  if (con) {
    printf("%c", con);
  }
}
long long N, M, K;
long long a[MAXN];
int t[MAXN][2];
int tot;
vector<int> pos[MAXN];
long long dp[30][2];
void insert(long long x, int idx) {
  int id = 0;
  int bit;
  for (int i = 29; i >= 0; i--) {
    bit = (x >> i) & 1;
    if (!t[id][bit]) {
      t[id][bit] = ++tot;
      t[tot][0] = t[tot][1] = 0;
      pos[tot].clear();
    }
    id = t[id][bit];
    pos[id].push_back(idx);
  }
}
void calc(int id, int level) {
  int l = t[id][0];
  int r = t[id][1];
  if (l) {
    calc(l, level - 1);
  }
  if (r) {
    calc(r, level - 1);
  }
  if (!l || !r) {
    return;
  }
  long long res = 0;
  int p = 0;
  for (auto x : pos[l]) {
    while (p < pos[r].size() && pos[r][p] < x) {
      p++;
    }
    res += p;
  }
  dp[level][0] += res;
  dp[level][1] += 1LL * pos[l].size() * pos[r].size() - res;
}
void init() {
  tot = 0;
  t[0][0] = t[0][1] = 0;
  for (int i = 0; i < 30; i++) {
    for (int j = 0; j < 2; j++) {
      dp[i][j] = 0;
    }
  }
}
void solve() {
  long long ans = 0;
  long long x = 0;
  N = read();
  init();
  for (int i = 1; i <= N; i++) {
    a[i] = read();
    insert(a[i], i);
  }
  calc(0, 29);
  for (int i = 0; i <= 29; i++) {
    if (dp[i][0] <= dp[i][1]) {
      ans += dp[i][0];
    } else {
      ans += dp[i][1];
      x += 1 << i;
    }
  }
  write(ans, ' ');
  write(x);
}
int main() {
  int T;
  T = 1;
  for (int t = 1; t <= T; t++) {
    GETDATA = false;
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).

An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.

You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.

Input

First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.

Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.

Examples

Input


4
0 1 3 2


Output


1 0


Input


9
10 7 9 10 7 5 5 3 5


Output


4 14


Input


3
8 10 3


Output


0 8

Note

In the first sample it is optimal to leave the array as it is by choosing x = 0.

In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:

  * i = 2, j = 3; 
  * i = 2, j = 4; 
  * i = 3, j = 4; 
  * i = 8, j = 9. 



In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,fma,tune=native")
using namespace std;
const int MAXN = 5e6 + 10;
const int MAXM = 2e6 + 10;
const long long MOD = 1e9 + 7;
const long long INF = 0x3f3f3f3f;
const unsigned long long BASE = 233;
const int base = 5;
mt19937_64 mt(time(0));
char rubbish[110];
bool GETDATA;
inline long long read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
inline void write(long long x, char con = '\n') {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10, 0);
  putchar(x % 10 + '0');
  if (con) {
    printf("%c", con);
  }
}
long long N, M, K;
long long a[MAXN];
int t[MAXN][2];
int tot;
vector<int> pos[MAXN];
long long dp[30][2];
void insert(long long x, int idx) {
  int id = 0;
  int bit;
  for (int i = 29; i >= 0; i--) {
    bit = (x >> i) & 1;
    if (!t[id][bit]) {
      t[id][bit] = ++tot;
      t[tot][0] = t[tot][1] = 0;
      pos[tot].clear();
    }
    id = t[id][bit];
    pos[id].push_back(idx);
  }
}
void calc(int id, int level) {
  int l = t[id][0];
  int r = t[id][1];
  if (l) {
    calc(l, level - 1);
  }
  if (r) {
    calc(r, level - 1);
  }
  if (!l || !r) {
    return;
  }
  long long res = 0;
  int p = 0;
  for (auto x : pos[l]) {
    while (p < pos[r].size() && pos[r][p] < x) {
      p++;
    }
    res += p;
  }
  dp[level][0] += res;
  dp[level][1] += 1LL * pos[l].size() * pos[r].size() - res;
}
void init() {
  tot = 0;
  t[0][0] = t[0][1] = 0;
  for (int i = 0; i < 30; i++) {
    for (int j = 0; j < 2; j++) {
      dp[i][j] = 0;
    }
  }
}
void solve() {
  long long ans = 0;
  long long x = 0;
  N = read();
  init();
  for (int i = 1; i <= N; i++) {
    a[i] = read();
    insert(a[i], i);
  }
  calc(0, 29);
  for (int i = 0; i <= 29; i++) {
    if (dp[i][0] <= dp[i][1]) {
      ans += dp[i][0];
    } else {
      ans += dp[i][1];
      x += 1 << i;
    }
  }
  write(ans, ' ');
  write(x);
}
int main() {
  int T;
  T = 1;
  for (int t = 1; t <= T; t++) {
    GETDATA = false;
    solve();
  }
  return 0;
}

```