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:64000000")
using namespace std;
const int inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
set<vector<int> > second;
int lp[1001000];
int ex[100100];
int q, r[1001000], n[1001000], ans[1001000];
int dp[22][1001000];
int calc(int a, int r) {
  if (r == 0) {
    if (a) return 2;
    return 1;
  }
  int &res = dp[a][r];
  if (res != -1) return res;
  res = 0;
  for (int i = 0; i <= a; i++) {
    res += calc(i, r - 1);
    if (res >= mod) res -= mod;
  }
  return res;
}
vector<int> get(int val) {
  vector<int> ans;
  while (val > 1) {
    int p = lp[val];
    int k = 0;
    while (val % p == 0) {
      k++;
      val /= p;
    }
    ans.push_back(k);
  }
  return ans;
}
int main() {
  for (int i = 2; i <= (int)1e6; i++) {
    if (lp[i] == 0) {
      for (int j = i; j <= (int)1e6; j += i) {
        if (lp[j] == 0) lp[j] = i;
      }
    }
  }
  memset(dp, -1, sizeof dp);
  ex[0] = 1;
  for (int i = 1; i < 100100; i++) {
    ex[i] = (ex[i - 1] + ex[i - 1]);
    if (ex[i] >= mod) ex[i] -= mod;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", r + i, n + i);
    int ans = 1;
    vector<int> v = get(n[i]);
    for (int j = 0; j < v.size(); j++) {
      ans = 1ll * ans * calc(v[j], r[i]) % mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}