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 MAXN = 1000001;
const int MAXFACT = MAXN + 100;
const int MOD = 1000000007;
int sito[MAXN];
long long fact[MAXFACT];
long long rfact[MAXFACT];
long long rev[MAXFACT];
int dp[8][8];
long long cnk(long long n, long long k) {
  if (k == 0) return 1;
  if (n < k) return 0;
  return fact[n] * rfact[k] % MOD * rfact[n - k] % MOD;
}
int main() {
  rev[1] = 1;
  for (int i = 2; i < MAXFACT; i++)
    rev[i] = (MOD - MOD / i * rev[MOD % i] % MOD) % MOD;
  fact[0] = 1;
  rfact[0] = 1;
  for (int i = 1; i < MAXFACT; i++) {
    fact[i] = fact[i - 1] * i % MOD;
    rfact[i] = rfact[i - 1] * rev[i] % MOD;
  }
  for (int i = 2; i * i < MAXN; i++)
    if (!sito[i]) {
      for (int j = i * i; j < MAXN; j += i) sito[j] = i;
    }
  for (int i = 2; i < MAXN; i++)
    if (!sito[i]) sito[i] = i;
  int q;
  vector<int> pw;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    pw.clear();
    while (n > 1) {
      int p = sito[n];
      int c = 0;
      while (n % p == 0) {
        n /= p;
        c++;
      }
      pw.push_back(c);
    }
    int k = pw.size();
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= i; j++) dp[i][j] = 0;
    dp[0][0] = 1;
    for (int i = 0; i < k; i++) {
      long long cnk1 = cnk(r + pw[i], pw[i]);
      long long cnk2 = cnk(r + pw[i] - 1, pw[i]);
      long long incv = (cnk1 - cnk2 + MOD) % MOD;
      for (int j = 0; j <= i; j++) {
        dp[i + 1][j] = (dp[i + 1][j] + dp[i][j] * cnk2) % MOD;
        dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j] * incv) % MOD;
      }
    }
    long long res = 0;
    for (int i = 0; i <= k; i++) res = (res + (1LL << i) * dp[k][i]) % MOD;
    printf("%lld\n", res);
  }
  cerr << clock() / double(CLOCKS_PER_SEC) << endl;
  return 0;
}