Problem: You are given n integers a1, a2, ..., an. Denote this list of integers as T.

Let f(L) be a function that takes in a non-empty list of integers L.

The function will output another integer as follows: 

  * First, all integers in L are padded with leading zeros so they are all the same length as the maximum length number in L. 
  * We will construct a string where the i-th character is the minimum of the i-th character in padded input numbers. 
  * The output is the number representing the string interpreted in base 10. 



For example f(10, 9) = 0, f(123, 321) = 121, f(530, 932, 81) = 30.

Define the function 

<image> Here, <image> denotes a subsequence.

In other words, G(x) is the sum of squares of sum of elements of nonempty subsequences of T that evaluate to x when plugged into f modulo 1 000 000 007, then multiplied by x. The last multiplication is not modded. 

You would like to compute G(0), G(1), ..., G(999 999). To reduce the output size, print the value <image>, where <image> denotes the bitwise XOR operator.

Input

The first line contains the integer n (1 ≤ n ≤ 1 000 000) — the size of list T.

The next line contains n space-separated integers, a1, a2, ..., an (0 ≤ ai ≤ 999 999) — the elements of the list. 

Output

Output a single integer, the answer to the problem.

Examples

Input

3
123 321 555


Output

292711924


Input

1
999999


Output

997992010006992


Input

10
1 1 1 1 1 1 1 1 1 1


Output

28160

Note

For the first sample, the nonzero values of G are G(121) = 144 611 577, G(123) = 58 401 999, G(321) = 279 403 857, G(555) = 170 953 875. The bitwise XOR of these numbers is equal to 292 711 924.

For example, <image>, since the subsequences [123] and [123, 555] evaluate to 123 when plugged into f.

For the second sample, we have <image>

For the last sample, we have <image>, where <image> is the binomial coefficient.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long INF = 1e9;
const int N = 1000031;
int n;
long long cnt[N];
long long SUM2[N], SUM[N];
long long pw2[N], pw10[N];
vector<int> val_list;
long long get_dig(int val, int ps) {
  val %= pw10[ps + 1];
  return val / pw10[ps];
}
int pcount(int x) {
  if (x == 0) return 0;
  return pcount(x / 2) + x % 2;
}
bool is_over(int a, int b) {
  for (int i = 0; i < 6; i++) {
    if (get_dig(a, i) < get_dig(b, i)) return false;
  }
  return true;
}
int cnt_over(int val) {
  return cnt[val];
  int res = 0;
  for (int i = 0; i < val_list.size(); i++) {
    if (is_over(val_list[i], val)) ++res;
  }
  return res;
}
long long get_squares_above(int val) {
  return SUM2[val];
  long long res = 0;
  for (int i = 0; i < val_list.size(); i++) {
    if (is_over(val_list[i], val)) res += 1ll * val_list[i] * val_list[i];
    res %= 1000000007;
  }
  return res % 1000000007;
}
long long get_s_above(int val) {
  return SUM[val];
  long long res = 0;
  for (int i = 0; i < val_list.size(); i++) {
    if (is_over(val_list[i], val)) res += 1ll * val_list[i];
    res %= 1000000007;
  }
  return res % 1000000007;
}
void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
int PC[2000];
int PREC_ADD[N];
int main() {
  for (int i = 0; i < 100; i++) {
    PC[i] = pcount(i);
  }
  pw2[0] = 1;
  for (int i = 1; i < N; i++) {
    pw2[i] = pw2[i - 1] * 2 % 1000000007;
  }
  pw10[0] = 1;
  for (int i = 1; i < 10; i++) {
    pw10[i] = pw10[i - 1] * 10 % 1000000007;
  }
  for (int i = 0; i < 64; i++) {
    int s = 0;
    for (int j = 0; j < 6; j++) {
      if (i & (1 << j)) s += pw10[j];
    }
    PREC_ADD[i] = s;
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int val;
    scanf("%d", &val);
    val_list.push_back(val);
    cnt[val]++;
  }
  for (int i = 0; i < 1000000; i++) {
    SUM[i] = cnt[i] * 1ll * i % 1000000007;
    SUM2[i] = cnt[i] * 1ll * i % 1000000007 * i % 1000000007;
  }
  for (int ps = 0; ps < 6; ps++) {
    for (int i = 0; i < 1000000; i++) {
      int here = get_dig(i, ps);
      for (int q = 0; q < here; q++) {
        int val2 = i - pw10[ps] * here + pw10[ps] * q;
        add(SUM[val2], SUM[i]);
        add(SUM2[val2], SUM2[i]);
        add(cnt[val2], cnt[i]);
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < 1000000; i++) {
    long long res = 0;
    if (cnt_over(i) == 0) continue;
    int bad_mask = 0;
    for (int j = 0; j < 6; j++) {
      if (get_dig(i, j) == 9) bad_mask |= (1 << j);
    }
    for (int mask = 0; mask < (1 << 6); mask++) {
      int here = i;
      int ok = 1;
      if (mask & bad_mask) ok = 0;
      here += PREC_ADD[mask];
      if (ok == 0) continue;
      long long cnt_above = cnt_over(here);
      if (cnt_above == 0) continue;
      long long s_squares_above = get_squares_above(here);
      long long s_above = get_s_above(here);
      long long addend_here = 0;
      if (cnt_above == 1) {
        addend_here = s_squares_above;
      } else {
        long long ways = pw2[cnt_above - 2];
        addend_here = ways * 1ll *
                      (s_squares_above + s_above * s_above % 1000000007) %
                      1000000007;
      }
      if (PC[mask] % 2 == 0)
        res += addend_here;
      else
        res -= addend_here;
      if (res < 0) res += 1000000007;
      if (res >= 1000000007) res -= 1000000007;
    }
    if (res != 0) {
      ans ^= (1ll * res * i);
    }
  }
  cout << ans << endl;
  cin.get();
  cin.get();
  return 0;
}