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;
template <class _Tp>
void ckmax(_Tp &a, _Tp b) {
  if (a < b) a = b;
}
template <class _Tp>
void ckmin(_Tp &a, _Tp b) {
  if (a > b) a = b;
}
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
  return (b == 0) ? (a) : (gcd(b, a % b));
}
int read() {
  char ch = getchar();
  bool f = 1;
  int x = 0;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') f = 0, ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = getchar();
  }
  return f ? x : -x;
}
const int inf = 1000000000;
const long long Inf = 1000000000000000000ll;
const long long mod = 1000000007;
bool vis[1000050];
int p[1000050], minp[1000050], cnt;
int f[1000050][22], sum[1000050][22];
int main() {
  for (int i = 2; i < 1000050; ++i) {
    if (!vis[i]) {
      p[++cnt] = i;
      minp[i] = i;
    }
    for (int j = 1; j <= cnt && i * p[j] < 1000050; ++j) {
      vis[i * p[j]] = 1;
      minp[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  f[0][0] = 1;
  sum[0][0] = 1;
  for (int i = 1; i < 22; ++i) {
    f[0][i] = 2;
    sum[0][i] = i * 2 + 1;
  }
  for (int i = 1; i < 1000050; ++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;
  }
  int T = read();
  while (T--) {
    int r = read(), n = read();
    int ans = 1;
    int tot;
    while (n > 1) {
      int t = minp[n];
      tot = 0;
      while (minp[n] == t) n /= t, ++tot;
      ans = (1ll * ans * f[r][tot]) % mod;
    }
    printf("%d\n", ans);
  }
  return 0;
}