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 <typename T>
inline void read(T &x) {
  register char ch;
  x = 0;
  while (!isdigit(ch = getchar()))
    ;
  do {
    x = (x << 1) + (x << 3) + ch - '0';
  } while (isdigit(ch = getchar()));
}
const int N = 1e6 + 5;
const long long mod = 1e9 + 7;
pair<int, short> con[N][8];
short ln[N];
int dp[2][30];
vector<int> que[N];
int ans[N], n[N];
int pri[N / 5];
bitset<N> notp;
inline void Init(int n) {
  int tot = 0;
  for (int i = 2, t; i <= n; i++) {
    if (!notp[i]) {
      con[i][ln[i]++] = make_pair(i, 1);
      pri[++tot] = i;
    }
    for (int j = 1; j <= tot; j++) {
      if ((t = i * pri[j]) > n) break;
      memcpy(con[t], con[i], sizeof con[i]);
      ln[t] = ln[i];
      notp[t] = 1;
      if (i % pri[j])
        con[t][ln[t]++] = make_pair(pri[j], 1);
      else {
        for (register short k = 0; k < ln[t]; k++)
          if (con[t][k].first == pri[j]) {
            con[t][k].second++;
            break;
          }
        break;
      }
    }
  }
}
int main() {
  int q, MaxR = 0, MaxN = 0;
  read(q);
  for (int i = 1, r, x; i <= q; i++) {
    read(r);
    read(n[i]);
    que[r].push_back(i);
    MaxR = max(MaxR, r);
    MaxN = max(MaxN, n[i]);
  }
  Init(MaxN);
  int cur = 1, lst = 0, nn, tmp;
  nn = 30;
  for (int i = 0; i <= MaxR; i++) {
    cur = lst ^ 1;
    dp[cur][0] = tmp = 1;
    if (i == 0)
      for (register short j = 1; j < 20; j++) dp[cur][j] = 2;
    else
      for (register short j = 1; j < 20; j++) {
        tmp += dp[lst][j];
        if (tmp >= mod) tmp -= mod;
        dp[cur][j] = tmp;
      }
    for (auto x : que[i]) {
      tmp = 1;
      nn = n[x];
      for (register short j = 0; j < ln[nn]; j++)
        tmp = ((long long)tmp * dp[cur][con[nn][j].second]) % mod;
      ans[x] = tmp;
    }
    lst = cur;
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}