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;
int const xn = 1e6 + 5, mod = 1e9 + 7;
int q, r, n, dp[xn][20], mnp[xn];
int rd() {
  int ret = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    ret = (ret << 3) + (ret << 1) + ch - '0', ch = getchar();
  return f ? ret : -ret;
}
int upt(long long x) {
  while (x >= mod) x -= mod;
  while (x < 0) x += mod;
  return x;
}
void init() {
  int mx = 1e6;
  mnp[1] = 1;
  for (int i = 2; i <= mx; i++)
    if (!mnp[i])
      for (int j = i; j <= mx; j += i) mnp[j] = i;
  dp[0][0] = 1;
  for (int i = 1; i <= 20; i++) dp[0][i] = 2;
  for (int i = 1, s = 0; i <= mx; i++, s = 0)
    for (int j = 0; j <= 20; j++) s = upt(s + dp[i - 1][j]), dp[i][j] = s;
}
int div(int x) {
  int ans = 1;
  while (x != 1) {
    int i = mnp[x], cnt = 0;
    while (x % i == 0) cnt++, x /= i;
    ans = ((long long)ans * dp[r][cnt]) % mod;
  }
  return ans;
}
int main() {
  q = rd();
  init();
  while (q--) {
    r = rd();
    n = rd();
    printf("%d\n", div(n));
  }
  return 0;
}