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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo(x, mod - 2); }
bool isPal(string ss) {
  int len = ss.length();
  for (int i = 0; i < len / 2; i++) {
    int comp = len - i - 1;
    if (ss[i] != ss[comp]) return false;
  }
  return true;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 1000001;
long long dp[25][M];
long long p[M], lo[M];
void sieve() {
  for (int i = 2; i < M; i++) {
    if (p[i]) continue;
    lo[i] = i;
    for (long long j = i * 1LL * i; j < M; j += i) {
      lo[j] = i;
      p[j] = 1;
    }
  }
}
void compute() {
  dp[0][0] = 1;
  for (int j = 1; j < 23; j++) {
    dp[j][0] = 2;
  }
  for (int i = 1; i < M; i++) {
    dp[0][i] = 1;
    for (int j = 1; j < 22; j++) {
      dp[j][i] = dp[j - 1][i] + dp[j][i - 1];
      if (dp[j][i] >= mod) dp[j][i] %= mod;
    }
  }
}
long long ans = 1;
long long r;
void dofor(long long x) {
  while (x > 1) {
    long long temp = lo[x];
    int times = 0;
    while (x % temp == 0) {
      x /= temp;
      times++;
    }
    ans *= dp[times][r];
    if (ans >= mod) ans %= mod;
  }
}
int main() {
  sieve();
  compute();
  int q;
  scanf("%d", &(q));
  while (q--) {
    long long n;
    scanf("%lld", &(r));
    scanf("%lld", &(n));
    ans = 1;
    dofor(n);
    printf("%lld\n", ans);
  }
}