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;
long long result[1000001][21], t[2][21], d[1000001], done[1000001],
    tsnt[1000001], mod = 1e9 + 7;
vector<long long> prime;
void dp() {
  t[0][0] = result[0][0] = 1;
  for (long long i = 1; i <= 20; i++)
    t[0][i] = t[0][i - 1] + 2, result[0][i] = 2;
  for (long long i = 1; i <= 1000000; i++)
    for (long long e = 0; e <= 20; e++) {
      result[i][e] = t[(i + 1) % 2][e];
      t[i % 2][e] = result[i][e];
      if (e != 0) t[i % 2][e] += t[i % 2][e - 1];
      t[i % 2][e] %= mod;
    }
}
void era() {
  for (long long i = 2; i <= 1000000; i++) {
    if (d[i] == 0) {
      for (long long e = 2; e <= 1000000; e++) {
        if (e * i > 1000000) break;
        d[i * e] = i;
      }
      d[i] = i;
    }
  }
}
void ptich(long long g, long long t) {
  long long u = g;
  prime.clear();
  while (u != 1) {
    if (done[d[u]] != t)
      done[d[u]] = t, tsnt[d[u]] = 1, prime.push_back(d[u]);
    else
      tsnt[d[u]]++;
    u /= d[u];
  }
}
long long get(long long g, long long level) {
  long long kq = 1;
  for (auto v : prime) kq *= result[level][tsnt[v]], kq %= mod;
  return kq;
}
void do_it() {
  dp();
  era();
  long long q, u, v, d = 0;
  cin >> q;
  while (q--) {
    d++;
    cin >> u >> v;
    ptich(v, d);
    cout << get(v, u) << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  do_it();
}