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 N = (int)1e6 + 6, mod = (int)1e9 + 7;
int pw(int a, int b) {
  return b <= 0 ? 1
                : pw(a * 1ll * a % mod, b >> 1) * 1ll * (b & 1 ? a : 1) % mod;
}
int f[N], inv[N], comb[N][25], dp[22][N], pr[N];
void sadd(int &x, int y) {
  x += y;
  while (x >= mod) x -= mod;
}
int32_t main() {
  for (int j = 0; j < N; ++j)
    for (int i = 0; i < 25; ++i)
      comb[j][i] =
          (j == i || !i) ? 1 : (comb[j - 1][i - 1] + comb[j - 1][i]) % mod;
  for (int j = 1; j < N; ++j) dp[0][j] = 1;
  for (int j = 2; j < N; ++j) {
    if (pr[j] == 0) {
      for (int i = j; i < N; i += j) {
        pr[i] = 1;
        dp[0][i] <<= 1;
      }
    }
  }
  for (int j = 1; j < 22; ++j) {
    for (int a = 1; a < N; ++a)
      if (dp[j - 1][a])
        for (int b = a + a; b < N; b += a) {
          sadd(dp[j][b], dp[j - 1][a]);
        }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int res = 0;
    for (int j = 0; j < 22; ++j) {
      if (j > r) continue;
      int ways = comb[r][j];
      res = (res + ways * 1ll * dp[j][n]) % mod;
    }
    printf("%d\n", res);
  }
}