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;
inline int readInt() {
  static int n, ch;
  n = 0, ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) n = n * 10 + ch - '0', ch = getchar();
  return n;
}
const int MAX_N = 1000000 + 3, MAX_K = 21, MOD = 1000000007;
int f[MAX_N][MAX_K];
int primes[MAX_N], mn[MAX_N], cnt = 0;
bool isPrime[MAX_N];
void prepare() {
  f[0][0] = 1;
  for (int i = 1; i < MAX_K; ++i) f[0][i] = 2;
  for (int i = 1; i < MAX_N; ++i) {
    f[i][0] = 1;
    for (int j = 1; j < MAX_K; ++j) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % MOD;
  }
  fill(isPrime + 2, isPrime + MAX_N, 1);
  for (int i = 2; i < MAX_N; ++i) {
    if (isPrime[i]) primes[cnt++] = i, mn[i] = i;
    for (int j = 0; j < cnt && i * primes[j] < MAX_N; ++j) {
      int t = i * primes[j];
      isPrime[t] = false;
      mn[t] = primes[j];
      if (i % primes[j] == 0) break;
    }
  }
}
int main() {
  prepare();
  int q = readInt();
  while (q--) {
    int r = readInt(), n = readInt(), t = n;
    long long ans = 1;
    while (n > 1) {
      int cc = 0, _mn = mn[n];
      while (n % _mn == 0) ++cc, n /= _mn;
      (ans *= f[r][cc]) %= MOD;
    }
    cout << ans << endl;
  }
  return 0;
}