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:

<image>

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 x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
bool b[1000000 + 5];
int Q, last[1000000 + 5], s[1000000 / 5 + 5], num, p[1000000 * 2 + 5],
    inv[1000000 * 2 + 5], a[65], c[65], ans, f[8][8], sum[25];
inline int C(int n, int m) {
  return m > n || m < 0
             ? 0
             : 1LL * p[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
inline int S(int n, int m) { return m ? C(n + m - 1, m - 1) : (n ? 0 : 1); }
int main() {
  p[0] = p[1] = inv[0] = inv[1] = sum[0] = 1;
  for (int i = 2; i <= 1000000; ++i) {
    if (!b[i]) s[++num] = i, last[i] = i;
    for (int j = 1; s[j] * i <= 1000000; ++j) {
      b[s[j] * i] = 1;
      last[s[j] * i] = s[j];
      if (i % s[j] == 0) break;
    }
  }
  for (int i = 2; i <= 1000000 * 2; ++i)
    p[i] = 1LL * p[i - 1] * i % 1000000007,
    inv[i] =
        1LL * (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  for (int i = 2; i <= 1000000 * 2; ++i)
    inv[i] = 1LL * inv[i - 1] * inv[i] % 1000000007;
  for (Q = read(); Q--;) {
    int m = read(), n = read();
    num = ans = 0;
    while (n > 1) {
      ++num;
      c[num] = 0;
      a[num] = last[n];
      for (int t = last[n]; n % t == 0; n /= t) ++c[num];
    }
    for (int i = 1; i <= 20; ++i) sum[i] = (sum[i - 1] + S(i, m)) % 1000000007;
    memset(f, 0, sizeof(f));
    f[0][0] = 1;
    for (int i = 1; i <= num; ++i)
      for (int j = 0; j < i; ++j) {
        f[i][j] = (f[i][j] + 1LL * f[i - 1][j] * S(c[i], m)) % 1000000007;
        f[i][j + 1] =
            (f[i][j + 1] + 1LL * f[i - 1][j] * sum[c[i] - 1]) % 1000000007;
      }
    for (int i = 0; i <= num; ++i)
      ans = (ans + 1LL * (1 << i) * f[num][i]) % 1000000007;
    printf("%d\n", ans);
  }
  return 0;
}