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>
using namespace std;
const long long mod = 1e9 + 7;
long long pp[2000001]{};
long long mypow(long long i, long long j) {
  if (pp[i]) return pp[i];
  auto ptr = &pp[i];
  long long ans = 1;
  for (ans = 1; j; j >>= 1, (i *= i) %= mod)
    if (j & 1) (ans *= i) %= mod;
  return *ptr = ans;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> p, f(k + 1, 0), mu(k + 1, 0);
  vector<vector<int>> num(k + 1);
  mu[1] = 1;
  for (int i = 2; i <= k; ++i) {
    if (!f[i]) p.push_back(i), f[i] = i;
    for (int j : p) {
      if (1ll * i * j > k) break;
      f[i * j] = j;
      if (i % j == 0) break;
    }
    if (i / f[i] % f[i]) mu[i] = -mu[i / f[i]];
  }
  for (int i = 1; i <= k; ++i)
    for (int j = i; mu[i] && j <= k; j += i) num[j].push_back(i);
  long long now = 0, ans = 0;
  for (int i = 1; i <= k; ++i) {
    for (int j : num[i])
      (now += (mypow(i / j, n) - mypow(i / j - 1, n)) * mu[j]) %= mod;
    (now += mod) %= mod;
    (ans += now ^ i) %= mod;
  }
  cout << ans;
}