Problem:
Score : 500 points

Problem Statement

Consider sequences \{A_1,...,A_N\} of length N consisting of integers between 1 and K (inclusive).

There are K^N such sequences. Find the sum of \gcd(A_1, ..., A_N) over all of them.

Since this sum can be enormous, print the value modulo (10^9+7).

Here \gcd(A_1, ..., A_N) denotes the greatest common divisor of A_1, ..., A_N.

Constraints

2 \leq N \leq 10^5

1 \leq K \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the sum of \gcd(A_1, ..., A_N) over all K^N sequences, modulo (10^9+7).

Sample Input 1

3 2

Sample Output 1

9

\gcd(1,1,1)+\gcd(1,1,2)+\gcd(1,2,1)+\gcd(1,2,2)
+\gcd(2,1,1)+\gcd(2,1,2)+\gcd(2,2,1)+\gcd(2,2,2)
=1+1+1+1+1+1+1+2=9

Thus, the answer is 9.

Sample Input 2

3 200

Sample Output 2

10813692

Sample Input 3

100000 100000

Sample Output 3

742202979

Be sure to print the sum modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;  // or pb_ds;

using namespace std;

#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#ifndef DEBUG
// #define endl "\n"
#endif
#define fff ff
#define sss ss.ff
#define ttt ss.ss
#define INF (1 << 30)
#define LLF (1ll << 60)
#define PRIME (1000000007)
#define FASTIO std::ios::sync_with_stdio(false)
#ifdef DEBUG
#define debug(x) cerr << #x << " = " << x << endl
#else
#define debug(x)
#endif
#define rep(i, begin, end)                            \
  for (__typeof(end) i = (begin) - ((begin) > (end)); \
       i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))

#define FILE_INPUT                  \
  freopen("input.txt", "r", stdin); \
  freopen("output.txt", "w", stdout)

#define all(x) x.begin(), x.end()

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef tuple<int, int, int> iii;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<string> vs;

template <typename T, typename B = null_type>
using oset =
    tree<T, B, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order / order_of_key

const ld eps = 1e-12;

/*-----------------END TEMPLATE-----------------*/

ll fastexp(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % PRIME;

    a = (a * a) % PRIME;
    b >>= 1;
  }
  return res;
}

void solve() {
  int n, k;
  cin >> n >> k;

  int ans[k + 1];
  ll sum = 0;
  for (int i = k; i >= 1; i--) {
    int divisors = k / i;
    ll quantosComI = fastexp(divisors, n);
    for (int j = i + i; j <= k; j += i) {
      quantosComI = (quantosComI - ans[j] + PRIME) % PRIME;
    }
    ans[i] = quantosComI;
    sum += ((ll)i * ans[i]) % PRIME;
    sum %= PRIME;
  }
  cout << sum << endl;
}

int(((((((((((((((main)))))))))))))))() {
  FASTIO;
  int n;
  // cin >> n;
  n = 1;
  int i = 1;
  while (n--) {
    solve();
  }
}

/*


00100
00

*/