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 <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& _p) {
  return os << "(" << _p.first << "," << _p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& _V) {
  bool f = true;
  os << "[";
  for (auto v : _V) {
    os << (f ? "" : ",") << v;
    f = false;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& _S) {
  bool f = true;
  os << "(";
  for (auto s : _S) {
    os << (f ? "" : ",") << s;
    f = false;
  }
  return os << ")";
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const map<T, U>& _M) {
  return os << set<pair<T, U>>(_M.begin(), _M.end());
}
const signed long long INF = 1000000100;
const long double EPS = 1e-9;
int P[1000100];
void sito() {
  for (int(i) = (2); (i) <= (1000100); (i)++) {
    if (P[i] != 0) continue;
    P[i] = i;
    for (long long j = (long long)i * i; j < 1000100; j += i) {
      P[j] = i;
    }
  }
}
const int N = 1000010;
unsigned int ans[N][20];
const int mod = 1000000007;
const int odw2 = 500000004;
int main() {
  sito();
  for (int(i) = (0); (i) < (20); (i)++) ans[0][i] = 2;
  ans[0][0] = 1;
  for (int(i) = (1); (i) < (N); (i)++)
    for (int(j) = (0); (j) < (20); (j)++) {
      for (int(k) = (0); (k) <= (j / 2); (k)++) {
        ans[i][j] =
            (ans[i][j] + ((k != j - k) ? (ans[i - 1][j - k] + ans[i - 1][k])
                                       : ans[i - 1][k]));
        if (ans[i][j] > mod) ans[i][j] %= mod;
      }
    }
  int q;
  scanf("%d", &q);
  for (int(i) = (1); (i) <= (q); (i)++) {
    int a, b;
    scanf("%d%d", &a, &b);
    long long ans1 = 1;
    while (b != 1) {
      int p = P[b];
      int d = 0;
      while (b % p == 0) {
        b /= p;
        d++;
      }
      ans1 = (ans1 * ans[a][d]);
      if (ans1 > mod) ans1 %= mod;
    }
    printf("%lld\n", ans1);
  }
}