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 MAX = 1 << 20;
const int MOD = 1e9 + 7;
int fak[MAX], invfak[MAX];
int dp[MAX][21];
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mul(int a, int b) { return (int)(((long long)a * b) % MOD); }
int pot(int x, int e) {
  int r = 1;
  for (; e; x = mul(x, x), e /= 2)
    if (e & 1) r = mul(r, x);
  return r;
}
int inv(int x) { return pot(x, MOD - 2); }
int divv(int a, int b) { return mul(a, inv(b)); }
void genfak() {
  fak[0] = invfak[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fak[i] = mul(fak[i - 1], i);
    invfak[i] = inv(fak[i]);
  }
}
int povrh(int a, int b) {
  if (a < b || b < 0) return 0;
  return mul(fak[a], mul(invfak[b], invfak[a - b]));
}
int sito[MAX];
void gen_sito() {
  for (int i = (2); i < (MAX); i++)
    if (!sito[i])
      for (int j = i; j < MAX; j += i) sito[j] = i;
}
vector<int> potencije(int x) {
  vector<int> R;
  for (; x > 1;) {
    int tmp = sito[x], kol = 0;
    for (; !(x % tmp); x /= tmp, kol++)
      ;
    R.push_back(kol);
  }
  return R;
}
int main() {
  genfak();
  gen_sito();
  dp[0][0] = 1;
  for (int j = (1); j < (21); j++) dp[0][j] = 2;
  for (int i = (1); i < (MAX); i++)
    for (int j = 0; j < 21; j++)
      for (int k = 0; k < j + 1; k++) dp[i][j] = add(dp[i][j], dp[i - 1][k]);
  int brq;
  scanf("%d", &brq);
  for (; brq--;) {
    int x, r;
    scanf("%d%d", &r, &x);
    auto R = potencije(x);
    int rje = 1;
    for (int i = 0; i < (int)R.size(); i++) rje = mul(rje, dp[r][R[i]]);
    printf("%d\n", rje);
  }
  return 0;
}