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;
int jie[2000050], i, ni[2000050], j, g[1000050], su[1000050], m;
bool vis[1000050];
int qmi(int di, int zhi) {
  int ret = 1, x = di;
  while (zhi) {
    if (zhi & 1) ret = 1LL * ret * x % 1000000007;
    x = 1LL * x * x % 1000000007;
    zhi >>= 1;
  }
  return ret;
}
int main() {
  jie[0] = jie[1] = ni[1] = ni[0] = 1;
  for (i = 2; i <= 2000000; i++) jie[i] = 1LL * jie[i - 1] * i % 1000000007;
  for (i = 1; i <= 2000000; i++) ni[i] = qmi(jie[i], 1000000007 - 2);
  for (i = 2; i <= 1000000; i++) {
    if (!vis[i]) {
      g[i] = i;
      su[++m] = i;
    }
    for (j = 1; j <= m; j++) {
      if (i * su[j] > 1000000) break;
      if (i % su[j]) {
        g[i * su[j]] = su[j];
        vis[i * su[j]] = true;
      } else {
        g[i * su[j]] = su[j];
        vis[i * su[j]] = true;
        break;
      }
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    int r, x;
    scanf("%d%d", &r, &x);
    int ans = 1;
    while (x != 1) {
      int cnt = 0, psz = g[x];
      while (x % psz == 0) x /= psz, cnt++;
      ans = 1LL * ans *
            (0LL +
             2LL * (r + cnt >= cnt
                        ? ((1LL * ni[r + cnt - cnt] * ni[cnt] % 1000000007) *
                           jie[r + cnt] % 1000000007)
                        : 0) -
             (r + cnt - 1 >= cnt
                  ? ((1LL * ni[r + cnt - 1 - cnt] * ni[cnt] % 1000000007) *
                     jie[r + cnt - 1] % 1000000007)
                  : 0)) %
            1000000007;
    }
    printf("%d\n", (ans + 1000000007) % 1000000007);
  }
  return 0;
}