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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3fll;
const int MAX_N = 1000010;
const long long mod = (int)(1e9) + 7;
int prime_cnt;
int prime[MAX_N], vis[MAX_N];
void Seive() {
  memset(vis, 0, sizeof(vis));
  prime_cnt = 0;
  for (int i = 2; i < MAX_N; ++i) {
    if (!vis[i]) prime[prime_cnt++] = i;
    for (int j = 0; j < prime_cnt && 1ll * prime[j] * i < MAX_N; ++j) {
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
long long dp[MAX_N][22];
void init() {
  for (int i = 0; i < MAX_N; ++i) dp[i][0] = 1;
  for (int i = 1; i < 22; ++i) dp[0][i] = 2;
  for (int i = 1; i < MAX_N; ++i) {
    for (int j = 1; j < 22; ++j) {
      dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
      if (dp[i][j] >= mod) dp[i][j] -= mod;
    }
  }
}
inline long long solve(int r, int n) {
  int t = n;
  long long ret = 1;
  for (int i = 0; i < prime_cnt && prime[i] * prime[i] <= t; ++i) {
    if (t % prime[i]) continue;
    int cnt = 0;
    while (t % prime[i] == 0) cnt++, t /= prime[i];
    ret = ret * dp[r][cnt] % mod;
  }
  if (t > 1) ret = ret * dp[r][1] % mod;
  return ret;
}
int main() {
  Seive();
  init();
  int Q, r, n;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &r, &n);
    printf("%lld\n", solve(r, n));
  }
  return 0;
}