// #include <atcoder/all>
// using namespace atcoder;
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
#define popcnt(x) __builtin_popcount(x)
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pil = pair<int, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vector<int>>;
using vvll = vector<vector<ll>>;
// const int inf = 1e9;
const ll inf = 1e18;
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};

void chmin(double& a, double b) { a = min(a, b); }

// const int mod = 998244353;
const int mod = 1000000007;
struct mint {
  ll x;  // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

map<pil,ll> dp;

ll f(int n,ll k){
  if (n==1 | k==1) return 2;
  pil p=make_pair(n,k);
  if (dp.count(p)) return dp[p];
  ll res=f(n-1,k);
  if (k>=n) res+=f(n-1,k/n);
  dp[p]=res;
  return res;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  ll k;
  cin>>n>>k;
  cout<<f(n,k)-1<<endl;

  return 0;
}