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 = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
template <class T>
inline T isqrt(T k) {
  T r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
template <class T>
inline T icbrt(T k) {
  T r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return (int)fpow(a, p - 2, p); }
const int MAXN = 1000050 + 10;
int fl[MAXN];
int sp[MAXN];
void Sieve() {
  for (int i = 2; i < MAXN; i += 2) sp[i] = 2;
  for (int i = 3; i < MAXN; i += 2)
    if (!fl[i]) {
      sp[i] = i;
      for (int j = i; (long long)i * j < MAXN; j += 2)
        if (!fl[i * j]) {
          fl[i * j] = 1;
          sp[i * j] = i;
        }
    }
}
const int maxn = 1000050 + 10;
int r, n;
int c[21][maxn];
void solve() {
  Sieve();
  for (int i = (0); i < (maxn); i++) c[0][i] = 1;
  for (int i = (1); i < (21); i++)
    for (int j = (1); j < (maxn); j++) {
      c[i][j] = c[i][j - 1] + c[i - 1][j - 1];
      if (c[i][j] >= MOD) c[i][j] -= MOD;
    }
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &r, &n);
    map<int, int> hs;
    while (n > 1) {
      hs[sp[n]]++;
      n /= sp[n];
    }
    if (r == 0) {
      printf("%d\n", (int)fpow(2, int((hs).size())));
      continue;
    }
    int ans = 1;
    for (__typeof((hs).begin()) it = (hs).begin(); it != (hs).end(); it++) {
      ans = mult(ans, c[it->second][r + it->second] +
                          c[it->second - 1][r + it->second - 1]);
    }
    printf("%d\n", ans);
  }
}
int main() {
  if (fopen("in.txt", "r")) {
    freopen("in.txt", "r", stdin);
  }
  solve();
  cerr << "\nTime elapsed: " << 1000.0 * clock() / CLOCKS_PER_SEC << " ms\n";
  return 0;
}