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>
int dp[1000000 + 10][20 + 1], lp[1000000 + 10], prime[1000000], tot;
bool vis[1000000 + 10];
using namespace std;
const long long mod = 1e9 + 7;
void init() {
  for (int i = (2); i <= (1000000); ++i) {
    if (!vis[i]) {
      prime[tot++] = i;
      lp[i] = i;
    }
    for (int j = (0); j < (tot); ++j) {
      if (i * prime[j] > 1000000) break;
      vis[i * prime[j]] = true;
      lp[i * prime[j]] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  for (int i = (1); i <= (20); ++i) dp[0][i] = 2;
  for (int i = (1); i <= (1000000); ++i) {
    dp[i][0] = 1;
    for (int j = (1); j <= (20); ++j)
      dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % mod;
  }
}
long long calc(int n, int r) {
  long long ans = 1;
  while (n > 1) {
    int pr = lp[n], cnt = 0;
    while (n % pr == 0) ++cnt, n /= pr;
    (ans *= dp[r][cnt]) %= mod;
  }
  return ans;
}
int main() {
  init();
  int T;
  scanf("%d", &T);
  while (T--) {
    int n, r;
    scanf("%d%d", &r, &n);
    printf("%I64d\n", calc(n, r));
  }
  return 0;
}