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>
template <typename Tp>
Tp Max(const Tp &a, const Tp &b) {
  return a > b ? a : b;
}
template <typename Tp>
Tp Min(const Tp &a, const Tp &b) {
  return a < b ? a : b;
}
template <typename Tp>
void Read(Tp &x) {
  Tp in = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    in = in * 10 + ch - '0';
    ch = getchar();
  }
  x = in * f;
}
const int MOD = 1000000007;
const int SN = 1000000 + 10;
long long f[SN][22], sum[SN], prime[SN], cnt, n, r;
bool vis[SN];
void pre() {
  vis[1] = 1;
  for (int i = 2; i <= SN - 10; i++) {
    if (!vis[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && prime[j] * i <= SN - 10; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  f[0][0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= 20; i++)
    f[0][i] = 2, sum[i] = (sum[i - 1] + f[0][i]) % MOD;
  for (int i = 1; i <= SN - 8; i++) {
    for (int j = 0; j <= 20; j++) f[i][j] = sum[j];
    sum[0] = f[i][0];
    for (int j = 1; j <= 20; j++) sum[j] = (sum[j - 1] + f[i][j]) % MOD;
  }
}
long long Ask(long long n, long long r) {
  int x = sqrt(n + 0.5);
  long long ans = 1;
  for (int i = 1; i <= cnt && prime[i] <= x && n != 1; i++) {
    if (n % prime[i] == 0) {
      int tmp = 0;
      while (n % prime[i] == 0) tmp++, n /= prime[i];
      ans = (ans * f[r][tmp]) % MOD;
    }
  }
  if (n != 1) ans = (ans * f[r][1]) % MOD;
  return ans;
}
int main() {
  int times = 0;
  Read(times);
  pre();
  while (times--) {
    Read(r), Read(n);
    printf("%I64d\n", Ask(n, r));
  }
  return 0;
}