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 inf = 0x3f3f3f3f;
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 << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
int die = 0;
const double eps = 1e-9;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
int q;
int r, n;
int f[N][20];
int sum[N][20];
int prime[N], tot = 0, vis[N];
void pre() {
  for (int i = (2); i <= (1e6); i++) {
    if (!vis[i]) prime[++tot] = i;
    for (int j = (1); j <= (tot); j++) {
      int x = prime[j] * i;
      if (x > 1e6) break;
      vis[x] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
void init() {
  sum[0][0] = f[0][0] = 1;
  for (int i = (1); i <= (20); i++)
    f[0][i] = 2, sum[0][i] = sum[0][i - 1] + f[0][i];
  for (int r = (1); r <= (1e6); r++) {
    sum[r][0] = f[r][0] = 1;
    for (int k = (1); k <= (20); k++) {
      f[r][k] = sum[r - 1][k];
      sum[r][k] = (sum[r][k - 1] + f[r][k]) % mod;
    }
  }
}
signed main() {
  init(), pre();
  q = read();
  while (q--) {
    r = read(), n = read();
    long long ans = 1;
    int m = sqrt(n + 0.5);
    for (int i = (1); i <= (tot); i++) {
      if (prime[i] > n || prime[i] > m) break;
      int cnt = 0;
      while (n % prime[i] == 0) cnt++, n /= prime[i];
      ans = (ans * f[r][cnt]) % mod;
    }
    if (n > 1) ans = (ans * f[r][1]) % mod;
    printf("%lld\n", ans);
  }
}