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 stdmod = 1000000007;
int spfact[1001000];
void seive(int n) {
  spfact[0] = spfact[1] = 0;
  for (int i = 0; i <= n; ++i) {
    spfact[i] = i;
  }
  for (int i = 2; i * i <= n; ++i)
    if (spfact[i] == i) {
      for (int j = i + i; j <= n; j += i) {
        if (spfact[j] == j) spfact[j] = i;
      }
    }
}
void pfact(int n, vector<int>& res) {
  res.clear();
  while (n > 1) {
    int p = spfact[n];
    int pwr = 0;
    while (n % p == 0) {
      n /= p;
      pwr++;
    }
    res.push_back(pwr);
  }
}
int dp[1000001][21];
int mul(int a, int b) {
  long long res = (long long)a * b;
  return (res >= stdmod ? res % stdmod : res);
}
int main() {
  ios::sync_with_stdio(false);
  seive(1000200);
  dp[0][0] = 1;
  for (int e = 1; e <= 20; ++e) dp[0][e] = 2;
  for (int r = 1; r <= 1000000; ++r) {
    dp[r][0] = 1;
    for (int e = 1; e <= 20; ++e)
      dp[r][e] = (dp[r - 1][e] + dp[r][e - 1]) % stdmod;
  }
  int q;
  scanf("%d", &q);
  vector<int> exps;
  while (q-- > 0) {
    int n, r;
    scanf("%d%d", &r, &n);
    pfact(n, exps);
    long long res = 1;
    for (auto e : exps) res = (res * dp[r][e]) % stdmod;
    printf("%lld\n", res);
  }
  return 0;
}