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;
inline int read() {
  int s = 0, t = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') t = -1;
    c = getchar();
  }
  while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
  return s * t;
}
const int N = 1e6 + 55, P = 1e9 + 7;
inline void Min(double &x, double v) {
  if (x > v) x = v;
}
inline void M(long long &x) {
  if (x >= P) x -= P;
}
int r, n, p[N], g[N], q[N], num[N], tot = 0;
bool vis[N];
long long f[N][25];
inline void pre(int inf) {
  int t;
  for (register int i = 2; i <= inf; i++) {
    if (!vis[i]) p[++tot] = i, g[i] = i;
    for (register int j = 1; j <= tot && (t = p[j] * i) <= inf; j++) {
      vis[t] = 1;
      if (i % p[j] == 0) {
        g[t] = g[i];
        break;
      }
      g[t] = p[j];
    }
  }
  f[0][0] = 1;
  for (register int i = 1; i <= 20; i++) f[0][i] = 2;
  for (register int i = 1; i <= inf; i++) {
    f[i][0] = 1;
    for (register int j = 1; j <= 20; j++)
      M(f[i][j] = f[i - 1][j] + f[i][j - 1]);
  }
}
int main() {
  int T = read();
  pre(1e6);
  while (T--) {
    r = read(), n = read();
    if (n == 1) {
      puts("1");
      continue;
    }
    tot = 0;
    while (n != 1) {
      q[++tot] = g[n], num[tot] = 0;
      while (n % q[tot] == 0) n /= q[tot], num[tot]++;
    }
    long long ans = 1;
    for (register int j = 1; j <= tot; j++) ans = ans * f[r][num[j]] % P;
    printf("%I64d\n", ans);
  }
  return 0;
}