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:

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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;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "sz = " << vc.size() << endl << "[";
  for (const T& v : vc) o << v << ",";
  o << "]";
  return o;
}
const long long ccc = 1000;
bool prime[ccc + 1];
vector<long long> pr;
void makeprime() {
  long long i, j;
  for (i = 2; i <= ccc; i++) prime[i] = true;
  for (i = 2; i * i <= ccc; i++)
    if (prime[i])
      for (j = 2; j * i <= ccc; j++) prime[j * i] = false;
  for (i = 2; i <= ccc; i++)
    if (prime[i]) pr.push_back(i);
}
const int MX = 1000000;
vector<int> rs[MX + 1];
int ns[MX + 1];
int vs[20];
typedef pair<int, int> P;
vector<P> qs[MX + 1];
int ans[MX + 1];
int mod = 1e9 + 7;
void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  makeprime();
  for (int i = 1; i <= (int)(MX); i++) ns[i] = i;
  for (long long p : pr) {
    for (int a = p; a <= MX; a += p) {
      int r = 0;
      while (ns[a] % p == 0) ns[a] /= p, r++;
      rs[a].push_back(r);
    }
  }
  for (int i = 1; i <= (int)(MX); i++)
    if (ns[i] > 1) rs[i].push_back(1);
  int Q;
  cin >> Q;
  for (int i = 0; i < (int)(Q); i++) {
    int r, n;
    scanf("%d%d", &r, &n);
    qs[r].push_back(P(i, n));
  }
  for (int i = 0; i < (int)(MX + 1); i++) {
    if (i == 0) {
      vs[0] = 1;
      for (int j = 1; j <= (int)(19); j++) vs[j] = 2;
    } else {
      for (int j = 1; j <= (int)(19); j++) add(vs[j], vs[j - 1]);
    }
    for (P p : qs[i]) {
      int id = p.first, n = p.second;
      long long ret = 1;
      for (int r : rs[n]) ret = ret * vs[r] % mod;
      ans[id] = ret;
    }
  }
  for (int i = 0; i < (int)(Q); i++) printf("%d\n", ans[i]);
}