Problem: Let's call an array a of size n coprime iff gcd(a1, a2, ..., an) = 1, where gcd is the greatest common divisor of the arguments.

You are given two numbers n and k. For each i (1 ≤ i ≤ k) you have to determine the number of coprime arrays a of size n such that for every j (1 ≤ j ≤ n) 1 ≤ aj ≤ i. Since the answers can be very large, you have to calculate them modulo 109 + 7.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 2·106) — the size of the desired arrays and the maximum upper bound on elements, respectively.

Output

Since printing 2·106 numbers may take a lot of time, you have to output the answer in such a way:

Let bi be the number of coprime arrays with elements in range [1, i], taken modulo 109 + 7. You have to print <image>, taken modulo 109 + 7. Here <image> denotes bitwise xor operation (^ in C++ or Java, xor in Pascal).

Examples

Input

3 4


Output

82


Input

2000000 8


Output

339310063

Note

Explanation of the example:

Since the number of coprime arrays is large, we will list the arrays that are non-coprime, but contain only elements in range [1, i]:

For i = 1, the only array is coprime. b1 = 1.

For i = 2, array [2, 2, 2] is not coprime. b2 = 7.

For i = 3, arrays [2, 2, 2] and [3, 3, 3] are not coprime. b3 = 25.

For i = 4, arrays [2, 2, 2], [3, 3, 3], [2, 2, 4], [2, 4, 2], [2, 4, 4], [4, 2, 2], [4, 2, 4], [4, 4, 2] and [4, 4, 4] are not coprime. b4 = 55.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long long mod = 1e9 + 7;
using namespace std;
void setIO(const string& file_name) {
  freopen((file_name + ".in").c_str(), "r", stdin);
  freopen((file_name + ".out").c_str(), "w+", stdout);
}
long long m[2100000], low[2100000], vis[2100000], BIT[2100000], fac[2100000];
long long freq[2100000], p[2100000], n, k, ans[2100000], contr[2100000];
long long bin_pow(long long a, long long b) {
  long long ans = 1ll;
  for (long long i = 0; (1 << (i)) <= b; i++) {
    if (b & (1 << (i))) {
      ans = (((ans) % mod * (a) % mod) % mod);
    }
    a = (((a) % mod * (a) % mod) % mod);
  }
  return ans;
}
long long S(int k, long long A) {
  return (!k) ? A % mod : S(k - ((k) & (-(k))), A + BIT[k]);
}
void U(int k, long long val) {
  for (; k <= n; k += ((k) & (-(k)))) BIT[k] += val;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  low[1] = mod;
  m[1] = 1;
  for (int i = 2; i <= 2e6; i++) {
    if (vis[i]) continue;
    for (int j = i; j <= 2e6; j += i) {
      if (!vis[j]) low[j] = i;
      vis[j] = 1;
    }
  }
  for (int i = 2; i <= 2e6; i++) {
    if (low[i] == i)
      m[i] = -1;
    else if ((i / low[i]) % low[i] == 0)
      m[i] = 0;
    else
      m[i] = m[i / low[i]] * -1;
  }
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    p[i] = bin_pow((long long)i, n);
  }
  long long tot = 0ll;
  for (int i = 1; i <= k; i++) {
    ans[i] = ans[i - 1];
    int ind = 0;
    for (int j = i; j; j /= low[j]) {
      if ((j / low[j]) % low[j] != 0 && j != 1) {
        fac[ind++] = low[j];
      }
    }
    for (int b = 0; b < (1 << (ind)); b++) {
      int x = 1;
      for (int j = 0; j < ind; j++)
        if (b & (1 << (j))) x *= fac[j];
      ans[i] += mod - contr[x];
      contr[x] = ((m[x] * p[i / x]) + mod) % mod;
      ans[i] += mod + contr[x];
    }
    ans[i] %= mod;
    tot += ans[i] ^ (long long)i;
  }
  cout << tot % mod << '\n';
  return 0;
}