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>
#pragma comment(linker, "/STACK:32000000")
using namespace std;
inline bool read(int& val) { return scanf("%d", &val) != -1; }
inline bool read(long long& val) { return scanf("%I64d", &val) != -1; }
inline bool read(double& val) { return scanf("%lf", &val) != -1; }
inline bool read(char* val) { return scanf("%s", val) != -1; }
template <class T1, class T2>
inline bool read(T1& a, T2& b) {
  return read(a) && read(b);
}
template <class T1, class T2, class T3>
inline bool read(T1& a, T2& b, T3& c) {
  return read(a) && read(b) && read(c);
}
template <class T1, class T2, class T3, class T4>
inline bool read(T1& a, T2& b, T3& c, T4& d) {
  return read(a) && read(b) && read(c) && read(d);
}
template <class T1, class T2, class T3, class T4, class T5>
inline bool read(T1& a, T2& b, T3& c, T4& d, T5& e) {
  return read(a) && read(b) && read(c) && read(d) && read(e);
}
const int N = 1111111;
const long long mod = 1000000007;
long long lp[N];
long long dp[N][22];
long long f0[N];
int main() {
  int n = 1000001, m = 20;
  for (int i = 1, _ = (n); i <= _; ++i) lp[i] = i;
  for (int i = (2), _ = (n); i < _; ++i)
    if (lp[i] == i)
      for (int j = i; j < n; j += i) lp[j] = i;
  for (int j = 1, _ = (m); j <= _; ++j) dp[0][j] = 2;
  for (int i = 0, _ = (n + 1); i < _; ++i) dp[i][0] = 1;
  for (int i = 1, _ = (n); i <= _; ++i)
    for (int j = 1, _ = (m); j <= _; ++j)
      dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
  int q;
  read(q);
  for (int i = 0, _ = (q); i < _; ++i) {
    long long ans = 1;
    int a, b;
    read(a, b);
    while (b != 1) {
      int cnt = 0, p = lp[b];
      while (b % p == 0) b /= p, cnt++;
      ans = ans * dp[a][cnt] % mod;
    }
    printf("%d\n", (int)ans);
  }
  return 0;
}