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>
const double PI = acos(-1.0);
using namespace std;
const long long mod = 1000000007;
const long long MAXR = 1e6 + 5;
const long long MAXE = 25;
long long r, n;
long long f[MAXR][MAXE];
bool vis[MAXR];
vector<long long> prime;
void setP() {
  long long m = sqrt(MAXR);
  for (long long i = 2; i <= m; i++)
    if (!vis[i])
      for (long long j = i * i; j < MAXR; j += i) vis[j] = true;
  for (long long i = 2; i < MAXR; i++)
    if (!vis[i]) prime.push_back(i);
}
void setD(long long r) {
  for (long long ri = 0; ri <= r; ri++) f[ri][0] = 1;
  for (long long ei = 1; ei < MAXE; ei++) f[0][ei] = 2;
  for (long long ri = 1; ri <= r; ri++)
    for (long long ei = 1; ei < MAXE; ei++)
      f[ri][ei] = (f[ri][ei - 1] + f[ri - 1][ei]) % mod;
}
vector<long long> pf;
vector<long long> exps;
void setPE(long long x) {
  for (long long i = 0; i < prime.size(); i++) {
    long long pi = prime[i];
    if (x % pi == 0) {
      pf.push_back(pi);
      long long ei = 0;
      while (x % pi == 0) {
        x /= pi;
        ei++;
      }
      exps.push_back(ei);
    }
    if (prime[i] * prime[i] > x) break;
  }
  if (x > 1) pf.push_back(x), exps.push_back(1);
}
vector<long long> vr, vn;
int main() {
  long long q;
  setP();
  scanf("%lld", &q);
  long long mr = 0;
  vr.clear();
  vn.clear();
  for (long long i = 0; i < q; i++) {
    long long tr, tn;
    scanf("%lld%lld", &tr, &tn);
    vr.push_back(tr);
    vn.push_back(tn);
    mr = max(mr, vr[i]);
  }
  setD(mr + 3);
  for (long long i = 0; i < q; i++) {
    r = vr[i];
    n = vn[i];
    pf.clear();
    exps.clear();
    setPE(n);
    long long ans = 1;
    for (long long i = 0; i < exps.size(); i++)
      ans = (ans * f[r][exps[i]]) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}