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;
int minP[N + 1];
const int mod = 1e9 + 7;
int add(int x, int y) { return (x += y) >= mod ? x - mod : x; }
template <class... T>
int add(int x, T... y) {
  return add(x, add(y...));
}
int mul(int x, int y) { return 1LL * x * y % mod; }
template <class... T>
int mul(int x, T... y) {
  return mul(x, mul(y...));
}
vector<int> fastPrimeFactorization(int n) {
  vector<int> ret;
  while (n != 1) {
    int cnt = 0;
    int u = minP[n];
    while (n % u == 0) {
      cnt++;
      n /= u;
    }
    if (cnt > 0) {
      ret.push_back(cnt);
    }
  }
  return ret;
}
int main() {
  for (int i = 0; i <= N; i++) {
    minP[i] = i;
  }
  for (int i = 2; i <= N; i++) {
    if (minP[i] == i) {
      for (int j = i * 2; j <= N; j += i) {
        minP[j] = i;
      }
    }
  }
  static int dp[N + 1][22];
  dp[0][0] = 1;
  for (int i = 1; i < 22; i++) {
    dp[0][i] = 2;
  }
  for (int i = 1; i <= N; i++) {
    dp[i][0] = 1;
    for (int j = 1; j < 22; j++) {
      dp[i][j] = add(dp[i][j - 1], dp[i - 1][j]);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int r, n;
    scanf("%d %d", &r, &n);
    int ans = 1;
    for (int e : fastPrimeFactorization(n)) {
      ans = mul(ans, dp[r][e]);
    }
    printf("%d\n", ans);
  }
}