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 = 1000000007;
const int W = 1000000;
const int K = 20;
int f[1010000][22], p[1010000], psz, minf[1010000], q;
bool vis[1010000];
int read_int() {
  char c;
  int i;
  bool neg = false;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') neg = true, c = getchar();
  for (i = c - '0', c = getchar(); c >= '0' && c <= '9';
       i = i * 10 + c - '0', c = getchar())
    ;
  return neg ? -i : i;
}
void init() {
  f[0][0] = 1;
  for (int i = 1; i <= K; ++i) f[0][i] = 2;
  for (int i = 1; i <= W; ++i) {
    f[i][0] = 1;
    for (int j = 1; j <= K; ++j) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % mod;
  }
  for (int i = 2; i <= W; ++i) {
    if (!vis[i]) p[++psz] = i, minf[i] = i;
    for (int j = 1; j <= psz && i * p[j] <= W; ++j) {
      int d = i * p[j];
      vis[d] = true;
      minf[d] = minf[d] == 0 ? p[j] : min(minf[d], p[j]);
      if (i % p[j] == 0) break;
    }
  }
}
char output[12000000];
int size;
void print(int x) {
  char s[20];
  int l = 0;
  if (x == 0) {
    s[l++] = '0';
  }
  while (x > 0) s[l++] = x % 10 + '0', x /= 10;
  for (int i = l - 1; i >= 0; --i) output[size++] = s[i];
  output[size++] = '\n';
}
int main() {
  q = read_int();
  init();
  for (int i = 1; i <= q; ++i) {
    int r, n;
    r = read_int();
    n = read_int();
    int ans = 1, last = -1, cnt = 0;
    while (n != 1) {
      if (minf[n] != last)
        ans = 1ll * ans * f[r][cnt] % mod, last = minf[n], cnt = 0;
      ++cnt, n /= minf[n];
    }
    ans = 1ll * ans * f[r][cnt] % mod;
    print(ans);
  }
  output[size - 1] = '\0';
  puts(output);
  return 0;
}