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;
int q, r, n, minp[1000001], pri[1000001], cnt;
int f[1000001][22], sum[1000001][22];
bool vis[1000001];
const int mod = 1e9 + 7;
inline int rd() {
  int x = 0, f = 1;
  char c = ' ';
  while (c < '0' || c > '9') f = c == '-' ? -1 : 1, c = getchar();
  while (c <= '9' && c >= '0') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
inline void pre() {
  for (int i = 2; i < 1000001; i++) {
    if (!vis[i]) pri[++cnt] = i, minp[i] = i;
    for (int j = 1; j <= cnt && i * pri[j] < 1000001; j++) {
      vis[i * pri[j]] = 1;
      minp[i * pri[j]] = pri[j];
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  q = rd();
  pre();
  f[0][0] = sum[0][0] = 1;
  for (int i = 1; i < 22; i++) f[0][i] = 2, sum[0][i] = sum[0][i - 1] + f[0][i];
  for (int i = 1; i < 1000001; i++) {
    for (int j = 0; j < 22; j++) f[i][j] = sum[i - 1][j];
    sum[i][0] = f[i][0];
    for (int j = 1; j < 22; j++) sum[i][j] = (sum[i][j - 1] + f[i][j]) % mod;
  }
  while (q--) {
    r = rd();
    n = rd();
    long long ans = 1;
    while (n > 1) {
      int t = minp[n], j = 0;
      while (n % t == 0) n /= t, j++;
      (ans *= f[r][j]) %= mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}