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>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(bool);
void precalc();
clock_t start;
int testNumber = 1;
bool todo = true;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  cin >> t;
  int testNum = 1;
  while (t--) {
    solve(true);
    ++testNumber;
  }
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
const long long L = 22;
const long long C = 1000500;
int32_t dp[C][L];
const int32_t mod = 1000000007;
int32_t least_prime[C];
void precalc() {
  for (long long i = 2; i < C; ++i) {
    if (!least_prime[i]) {
      least_prime[i] = i;
      for (long long j = i * i; j < C; j += i) {
        least_prime[j] = i;
      }
    }
  }
  dp[0][0] = 1;
  for (long long i = 1; i < L; ++i) {
    dp[0][i] = 2 % mod;
  }
  for (long long deg = 1; deg < C; ++deg) {
    for (long long i = 0; i < L; ++i) {
      long long s = 0;
      for (long long j = 0; j <= i; ++j) {
        s += dp[deg - 1][j] + dp[deg - 1][i - j];
        if (s > 2 * mod) {
          s -= 2 * mod;
        }
      }
      s /= 2;
      dp[deg][i] = int32_t(s) % mod;
    }
  }
}
void solve(bool read) {
  long long r, n;
  cin >> r >> n;
  if (n == 1) {
    cout << 1 << "\n";
    return;
  }
  long long res = 1;
  while (n > 1) {
    long long cur = least_prime[n];
    long long cnt = 0;
    while (n % cur == 0) {
      n /= cur;
      ++cnt;
    }
    res = res * dp[r][cnt] % mod;
  }
  cout << res << "\n";
}