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 mod = 1e9 + 7;
const int N = 1e6 + 7;
int Q, tot;
int d[N], prime[N], f[N][21], num[N];
int sta[N], top;
bool check[N], had[N];
int qread() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
void calc(int x) {
  while (top) had[sta[top]] = num[sta[top--]] = 0;
  while (x != 1) {
    if (!had[d[x]]) {
      sta[++top] = d[x];
      had[d[x]] = 1;
    }
    num[d[x]]++;
    x /= d[x];
  }
}
int main() {
  scanf("%d", &Q);
  check[1] = d[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!check[i]) prime[++tot] = i, d[i] = i;
    for (int j = 1; j <= tot && prime[j] * i < N; j++) {
      check[prime[j] * i] = 1;
      d[prime[j] * i] = prime[j];
      if (i % prime[j] == 0) break;
    }
  }
  f[0][1] = 1;
  for (int i = 2; i <= 20; i++) f[0][i] = 2;
  for (int i = 1; i < N; i++)
    for (int j = 1; j <= 20; j++) f[i][j] = (f[i][j - 1] + f[i - 1][j]) % mod;
  int r, n;
  long long ans;
  while (Q--) {
    r = qread();
    n = qread();
    ans = 1;
    calc(n);
    for (int i = 1; i <= top; i++)
      ans = 1ll * ans * f[r][num[sta[i]] + 1] % mod;
    printf("%d\n", ans);
  }
  return 0;
}