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;
const int MAXN = 1000006;
const int LOGN = 20;
const int MOD = 1000000007;
int sieve[MAXN];
inline int sum(int a, int b) {
  int c = a + b;
  if (c >= MOD) {
    c -= MOD;
  }
  return c;
}
inline int prod(int a, int b) { return (1LL * a * b) % MOD; }
void gen_sieve() {
  memset(sieve, -1, sizeof(sieve));
  for (int i = 2; i < MAXN; i += 2) {
    sieve[i] = 2;
  }
  for (int i = 3; i < MAXN; i += 2) {
    if (sieve[i] != -1) {
      continue;
    }
    sieve[i] = i;
    for (long long j = 1LL * i * i; j < MAXN; j += 2 * i) {
      sieve[j] = i;
    }
  }
}
int table[MAXN][LOGN];
int main() {
  ios_base::sync_with_stdio(false);
  gen_sieve();
  table[0][0] = 1;
  for (int k = 1; k < LOGN; ++k) {
    table[0][k] = 2;
  }
  for (int r = 1; r < MAXN; ++r) {
    table[r][0] = 1;
    for (int k = 1; k < LOGN; ++k) {
      table[r][k] = sum(table[r][k - 1], table[r - 1][k]);
    }
  }
  int q;
  scanf("%d", &q);
  int r, n;
  while (q-- > 0) {
    scanf("%d %d", &r, &n);
    int ans = 1;
    while (n > 1) {
      int pf = sieve[n];
      int k = 0;
      while (n % pf == 0) {
        ++k;
        n /= pf;
      }
      ans = prod(ans, table[r][k]);
    }
    printf("%d\n", ans);
  }
  return 0;
}