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;
inline int getnum() {
  int r;
  scanf("%d", &r);
  return r;
}
inline void putnum(int x) { printf("%d", x); }
int n, q, dp[1000111][20];
int p[100111], pn, mxcnt[1000111], pre[1000111];
bool np[1000111];
int main() {
  dp[0][0] = 1;
  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];
      dp[i][j] = dp[i][j] >= mod ? dp[i][j] - mod : dp[i][j];
    }
  }
  np[1] = 1;
  for (int i = 2; i <= 1000000; i++) {
    if (!np[i]) {
      p[pn++] = i;
      pre[i] = 1;
      mxcnt[i] = 1;
    }
    for (int j = 0; j < pn && i * p[j] <= 1000000; j++) {
      np[i * p[j]] = 1;
      pre[i * p[j]] = pre[i] * p[j];
      mxcnt[i * p[j]] = mxcnt[i];
      if (i % p[j] == 0) {
        if (pre[i] == 1) {
          pre[i * p[j]] = 1;
          mxcnt[i * p[j]] = mxcnt[i] + 1;
        }
        break;
      }
    }
  }
  n = getnum();
  for (int i = 1; i <= n; i++) {
    int x = getnum(), y = getnum(), ans = 1;
    while (y > 1) {
      ans = 1ll * ans * dp[x][mxcnt[y]] % mod;
      y = pre[y];
    }
    putnum(ans);
    puts("");
  }
  return 0;
}