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;
const int mod = 1e9 + 7;
const int N = 2000100;
int vis[N], pri[N], is[N], tot = 0, n, m, q;
long long jc[N], inv[N], ans;
long long ksm(long long x, int y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
void ycl() {
  int i, j;
  for (i = 2; i <= 1000000; i++) {
    if (!vis[i]) {
      pri[++tot] = i;
      is[i] = i;
    }
    for (j = 1; ((i * pri[j] <= 1000000) && (j <= tot)); j++) {
      vis[i * pri[j]] = 1;
      is[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
  jc[0] = 1;
  for (i = 1; i <= 2000000; i++) jc[i] = (jc[i - 1] * (long long)i) % mod;
  inv[0] = 1;
  inv[2000000] = ksm(jc[2000000], mod - 2);
  for (i = 1999999; i; i--) inv[i] = (inv[i + 1] * (long long)(i + 1)) % mod;
}
long long C(int n, int m) {
  return ((jc[n] * inv[m]) % mod * inv[n - m]) % mod;
}
int main() {
  int i, j, k, x, cnt;
  long long tmp;
  ycl();
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &n, &m);
    ans = 1;
    while (m != 1) {
      x = is[m];
      cnt = 0;
      while (m % x == 0) cnt++, m /= x;
      if (!n)
        ans = (ans * 2) % mod;
      else {
        tmp = (2 * C(cnt + n, n)) % mod;
        tmp = (tmp - C(cnt + n - 1, n - 1) + mod) % mod;
        ans = (ans * tmp) % mod;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}