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;
bool isp[1000500];
int pr[1000500], g[1000500], F[1000500][20 + 1], n, r, k, ans, cur, cnt;
inline int rd() {
  int r;
  scanf("%d", &r);
  return r;
}
void _pre() {
  for (int _ = 2; _ <= 1000000; _++) {
    if (!isp[_]) pr[++cnt] = _;
    for (int i = 1; i <= cnt && pr[i] * _ <= 1000000; i++) {
      isp[_ * pr[i]] = 1;
      if (_ % pr[i] == 0) break;
    }
  }
  for (int _ = 2; _ <= 1000000; _++)
    if (isp[_]) {
      for (int i = 1; i <= cnt; i++)
        if (_ % pr[i] == 0) {
          g[_] = pr[i];
          break;
        }
    } else
      g[_] = _;
  for (int _ = 0; _ <= 1000000; _++) F[_][0] = 1;
  for (int _ = 1; _ <= 20; _++) F[0][_] = 2;
  for (int _ = 1; _ <= 1000000; _++)
    for (int i = 1; i <= 20; i++)
      F[_][i] = (F[_ - 1][i] + F[_][i - 1]) % 1000000007;
}
int main() {
  _pre();
  for (int T = rd(); T; T--) {
    r = rd(), n = rd(), ans = 1, cur = 0, k = 0;
    while (n != 1) {
      if (g[n] == cur) {
        k++;
      } else {
        ans = 1LL * ans * F[r][k] % 1000000007;
        cur = g[n];
        k = 1;
      }
      n /= g[n];
    }
    k ? ans = 1LL * ans * F[r][k] % 1000000007 : 0;
    printf("%d\n", (int)ans);
  }
  return 0;
}