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 = 1e6 + 1;
const int LOG = 21;
const int mod = 1e9 + 7;
int dp[N][LOG];
bool pr[N];
inline void add(int &x, int y) { x = (x + y) % mod; }
inline int sub(int x, int y) { return ((x - y) % mod + mod) % mod; }
void init() {
  dp[0][0] = 1;
  for (int i = 1; i < LOG; i++) dp[0][i] = 2, add(dp[0][i], dp[0][i - 1]);
  for (int i = 1; i < N; i++)
    for (int j = 0; j < LOG; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j) add(dp[i][j], dp[i][j - 1]);
    }
  memset(pr, 1, sizeof(pr));
  pr[0] = pr[1] = 0;
  for (int i = 2; i * i < N; i++) {
    if (pr[i]) {
      for (int j = i * i; j < N; j += i) pr[j] = 0;
    }
  }
}
int main() {
  init();
  int q;
  scanf("%d", &q);
  while (q--) {
    int r, n;
    scanf("%d%d", &r, &n);
    int ans = 1;
    for (int i = 2; i * i <= n; i++) {
      if (pr[i] && n % i == 0) {
        int cnt = 0;
        while (n % i == 0) cnt++, n /= i;
        ans = 1LL * ans * sub(dp[r][cnt], dp[r][cnt - 1]) % mod;
      }
    }
    if (n > 1) ans = 1LL * ans * sub(dp[r][1], dp[r][0]) % mod;
    printf("%d\n", ans);
  }
  return 0;
}