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;
string dst__;
stringstream dss__;
void debug(stringstream& ds) {
  cerr << endl;
  return;
}
template <typename F, typename... R>
void debug(stringstream& ds, const F& f, const R&... r) {
  string n;
  ds >> n;
  cerr << n << '=' << f << ' ';
  debug(ds, r...);
}
constexpr long long LINF = 334ll << 53;
constexpr int INF = 15 << 26;
constexpr long long MOD = 1E9 + 7;
void sieve(long long n, vector<long long>& p) {
  vector<char> isprime(n + 1, 1);
  isprime[0] = isprime[1] = 0;
  for (long long i = 2; i <= n; i++) {
    if (isprime[i]) {
      p.push_back(i);
      for (long long j = i * i; j <= n; j += i) isprime[j] = 0;
    }
  }
}
vector<long long> primes;
vector<vector<long long>> dp(22, vector<long long>(1000010));
long long r, c;
long long primeFactor(long long k) {
  vector<pair<long long, int>> ret;
  for (int i = 0;; ++i) {
    if (primes[i] * primes[i] > k) {
      if (k != 1ll) ret.push_back(make_pair(k, 1));
      break;
    }
    if (k % primes[i] == 0) {
      ret.push_back(make_pair(primes[i], 0));
      while (k % primes[i] == 0) {
        k /= primes[i];
        ret.back().second++;
      }
    }
  }
  long long ans = 1;
  for (auto& y : ret) {
    ans = ans * dp[y.second][r] % MOD;
  }
  return ans;
}
int main() {
  sieve(1000003, primes);
  for (int i = 1; i < 22; ++i) {
    dp[i][0] = 2;
    if (i == 1) {
      for (int j = 1; j < 1000010; ++j) {
        dp[i][j] = j + 2;
      }
    } else {
      for (int j = 1; j < 1000010; ++j) {
        dp[i][j] = dp[i - 1][j] + dp[i][j - 1] % MOD;
      }
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%lld%lld", &r, &c);
    auto ans = primeFactor(c);
    printf("%lld\n", ans);
  }
  return 0;
}