Problem: Bash got tired on his journey to become the greatest Pokemon master. So he decides to take a break and play with functions.

Bash defines a function f0(n), which denotes the number of ways of factoring n into two factors p and q such that gcd(p, q) = 1. In other words, f0(n) is the number of ordered pairs of positive integers (p, q) such that p·q = n and gcd(p, q) = 1.

But Bash felt that it was too easy to calculate this function. So he defined a series of functions, where fr + 1 is defined as:

<image>

Where (u, v) is any ordered pair of positive integers, they need not to be co-prime.

Now Bash wants to know the value of fr(n) for different r and n. Since the value could be huge, he would like to know the value modulo 109 + 7. Help him!

Input

The first line contains an integer q (1 ≤ q ≤ 106) — the number of values Bash wants to know.

Each of the next q lines contain two integers r and n (0 ≤ r ≤ 106, 1 ≤ n ≤ 106), which denote Bash wants to know the value fr(n).

Output

Print q integers. For each pair of r and n given, print fr(n) modulo 109 + 7 on a separate line.

Example

Input

5
0 30
1 25
3 65
2 5
4 48


Output

8
5
25
4
630

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr int MOD = 1000000007;
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.first) ^ (u(p.second) << 7);
  }
};
}  // namespace std
template <typename T, typename F>
T bsh(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      l = m + 1;
      r = m;
    } else {
      h = m - 1;
    }
  }
  return r;
}
template <typename T, typename F>
T bsl(T l, T h, const F& f) {
  T r = -1, m;
  while (l <= h) {
    m = (l + h) / 2;
    if (f(m)) {
      h = m - 1;
      r = m;
    } else {
      l = m + 1;
    }
  }
  return r;
}
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <unsigned int N>
class Field {
  inline unsigned int inv(unsigned int a) {
    long long t = 0, nt = 1, r = N, nr = a;
    while (nr) {
      long long q = r / nr;
      t -= q * nt;
      swap(t, nt);
      r -= q * nr;
      swap(r, nr);
    }
    assert(r <= 1);
    return (t < 0) ? t + N : t;
  }

 public:
  inline Field(int first = 0) : v(first) {}
  inline Field<N>& operator+=(const Field<N>& o) {
    if ((long long)v + o.v >= N)
      v = (long long)v + o.v - N;
    else
      v = v + o.v;
    return *this;
  }
  inline Field<N>& operator-=(const Field<N>& o) {
    if (v < o.v)
      v = N - o.v + v;
    else
      v = v - o.v;
    return *this;
  }
  inline Field<N>& operator*=(const Field<N>& o) {
    v = (unsigned long long)v * o.v % N;
    return *this;
  }
  inline Field<N>& operator/=(const Field<N>& o) { return *this *= inv(o.v); }
  inline Field<N> operator+(const Field<N>& o) {
    Field<N> r{*this};
    return r += o;
  }
  inline Field<N> operator-(const Field<N>& o) {
    Field<N> r{*this};
    return r -= o;
  }
  inline Field<N> operator*(const Field<N>& o) {
    Field<N> r{*this};
    return r *= o;
  }
  inline Field<N> operator/(const Field<N>& o) {
    Field<N> r{*this};
    return r /= o;
  }
  inline Field<N> operator-() { return {-v}; };
  inline Field<N>& operator++() {
    ++v;
    if (v == N) v = 0;
    return *this;
  }
  inline Field<N> operator++(int) {
    Field<N> r{*this};
    ++*this;
    return r;
  }
  inline Field<N>& operator--() {
    --v;
    if (v == -1) v = N - 1;
    return *this;
  }
  inline Field<N> operator--(int) {
    Field<N> r{*this};
    --*this;
    return r;
  }
  inline bool operator==(const Field<N>& o) { return o.v == v; }
  inline explicit operator unsigned int() const { return v; }
  inline static vector<Field<N>> fact(int t) {
    vector<Field<N>> F(t + 1, 1);
    for (int i = 2; i <= t; ++i) {
      F[i] = F[i - 1] * i;
    }
    return F;
  }
  inline static vector<Field<N>> invfact(int t) {
    vector<Field<N>> F(t + 1, 1);
    for (int i = 2; i <= t; ++i) {
      F[i] = F[i - 1] / i;
    }
    return F;
  }

 private:
  unsigned int v;
};
template <unsigned int N>
istream& operator>>(std::istream& is, Field<N>& f) {
  unsigned int v;
  is >> v;
  f = v;
  return is;
}
template <unsigned int N>
ostream& operator<<(std::ostream& os, const Field<N>& f) {
  return os << (unsigned int)f;
}
template <unsigned int N>
Field<N> operator+(int i, const Field<N>& f) {
  return Field<N>(i) + f;
}
template <unsigned int N>
Field<N> operator-(int i, const Field<N>& f) {
  return Field<N>(i) - f;
}
template <unsigned int N>
Field<N> operator*(int i, const Field<N>& f) {
  return Field<N>(i) * f;
}
template <unsigned int N>
Field<N> operator/(int i, const Field<N>& f) {
  return Field<N>(i) / f;
}
class TaskE {
 public:
  vector<int> F;
  void solve(istream& cin, ostream& cout) {
    cin.tie(nullptr);
    cout.tie(nullptr);
    F.resize(1000000 + 1);
    for (int i = 2; i <= 1000000; ++i) {
      if (F[i] == 0) {
        for (int j = i; j <= 1000000; j += i) {
          F[j] = i;
        }
      }
    }
    vector<vector<Field<1000000007>>> R(1000000 + 1,
                                        vector<Field<1000000007>>(20, 0));
    for (int k = 0; k < 20; ++k) {
      R[0][k] = k == 0 ? 1 : 2;
    }
    for (int l = 1; l <= 1000000; ++l) {
      R[l][0] = R[l - 1][0];
      for (int k = 1; k < 20; ++k) {
        R[l][k] = R[l - 1][k] + R[l][k - 1];
      }
    }
    int Q;
    cin >> Q;
    for (int q = 0; q < Q; ++q) {
      int r, n;
      cin >> r >> n;
      Field<1000000007> ans = 1;
      while (n > 1) {
        int f = F[n];
        int c = 0;
        while (n > 1 && F[n] == f) {
          n /= f;
          ++c;
        }
        ans *= R[r][c];
      }
      cout << ans << '\n';
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}