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 = 1e6 + 7;
const int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const int maxl = 20;
const int SQ = 400;
int dp[maxn][maxl];
int mark[maxn];
void prep() {
  for (int i = 2; i < maxn; i++) {
    if (mark[i]) continue;
    for (int j = i; j < maxn; j += i) mark[j] = i;
  }
  for (int i = 0; i < maxn; i++)
    for (int j = 0; j < maxl; j++)
      dp[i][j] =
          (i > 0 ? (dp[i - 1][j] + (j > 0 ? dp[i][j - 1] : 0)) % mod : 2 - !j);
}
int main() {
  prep();
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    long long ans = 1;
    for (int i = mark[n]; n > 1; i = mark[n]) {
      int cnt = 0;
      while (n % i == 0) cnt++, n /= i;
      ans = (1ll * ans * dp[r][cnt]) % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}