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;
const int MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int p[1000001], d[1000001][21];
int main() {
  for (int(i) = (2); (i) <= (1000000); (i) += (1)) {
    for (int(j) = (i); (j) <= (1000000); (j) += (i))
      if (!p[j]) p[j] = i;
  }
  for (int(r) = (0); (r) <= (1000000); (r) += (1)) d[r][0] = 1;
  for (int(i) = (1); (i) <= (20); (i) += (1)) d[0][i] = 2;
  for (int(i) = (1); (i) <= (1000000); (i) += (1))
    for (int(j) = (1); (j) <= (20); (j) += (1))
      d[i][j] = (d[i][j - 1] + d[i - 1][j]) % MOD;
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    long long ans = 1;
    vector<pair<int, int> > v;
    while (n > 1) {
      int x = p[n];
      if (!v.empty() && v.back().first == x)
        v.back().second++;
      else
        v.push_back(make_pair(x, 1));
      n /= x;
    }
    for (pair<int, int> p : v) ans = (ans * d[r][p.second]) % MOD;
    printf("%lld\n", ans);
  }
}