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 maxn = 1100000;
const int mod = 1000000007;
int q, n, r;
int s[maxn][25];
int p[maxn], prime[maxn];
int lp[maxn];
long long ans;
void work() {
  int i, j, k, last, sum;
  for (i = 0; i <= 1000000; i++)
    for (j = 0; j <= 24; j++) {
      if (i == 0)
        s[i][j] = 2;
      else {
        if (i != 1) {
          if (j != 0) {
            s[i][j] = (s[i - 1][j] + s[i][j - 1]);
            if (s[i][j] >= mod) s[i][j] -= mod;
          } else
            s[i][j] = 1;
        } else
          s[i][j] = j * 2 + 1;
      }
    }
  s[0][0] = 1;
  p[1] = 1;
  k = 0;
  for (i = 2; i <= 1000000; i++) {
    if (p[i] == 0) {
      lp[i] = i;
      prime[++k] = i;
    }
    for (j = 1; lp[i] >= prime[j] && j <= k && prime[j] * i <= 1000000; j++) {
      p[prime[j] * i] = 1;
      lp[prime[j] * i] = prime[j];
    }
  }
  k = 0;
  for (i = 2; i <= 1000000; i++)
    if (p[i] == 0) prime[++k] = i;
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d", &r, &n);
    ans = 1;
    sum = 0;
    last = -1;
    while (n != 1) {
      if (lp[n] != last) {
        if (sum != 0) ans = (ans * (long long)s[r][sum]) % mod;
        sum = 1;
        last = lp[n];
      } else {
        sum++;
      }
      n = n / lp[n];
    }
    ans = (ans * (long long)s[r][sum]) % mod;
    printf("%I64d\n", ans);
  }
}
int main() {
  work();
  return 0;
}