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:

<image>

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;
using ll = long long;
template <typename...>
static inline int getchar_unlocked(void) {
  return getchar();
}
template <typename...>
static inline void putchar_unlocked(int c) {
  putchar(c);
}
inline void reader(int& x) {
  int k, m = 0;
  x = 0;
  for (;;) {
    (k) = getchar_unlocked();
    if (k == '-') {
      m = 1;
      break;
    }
    if ('0' <= k && k <= '9') {
      x = k - '0';
      break;
    }
  }
  for (;;) {
    (k) = getchar_unlocked();
    if (k < '0' || k > '9') break;
    x = x * 10 + k - '0';
  }
  if (m) x = -x;
}
inline void reader(ll& x) {
  int k, m = 0;
  x = 0;
  for (;;) {
    (k) = getchar_unlocked();
    if (k == '-') {
      m = 1;
      break;
    }
    if ('0' <= k && k <= '9') {
      x = k - '0';
      break;
    }
  }
  for (;;) {
    (k) = getchar_unlocked();
    if (k < '0' || k > '9') break;
    x = x * 10 + k - '0';
  }
  if (m) x = -x;
}
inline int reader(char c[]) {
  int i, s = 0;
  for (;;) {
    (i) = getchar_unlocked();
    if (i != ' ' && i != '\n' && i != '\r' && i != '\t' && i != EOF) break;
  }
  c[s++] = i;
  for (;;) {
    (i) = getchar_unlocked();
    if (i == ' ' || i == '\n' || i == '\r' || i == '\t' || i == EOF) break;
    c[s++] = i;
  }
  c[s] = '\0';
  return s;
}
inline int reader(string& c) {
  int i;
  for (;;) {
    (i) = getchar_unlocked();
    if (i != ' ' && i != '\n' && i != '\r' && i != '\t' && i != EOF) break;
  }
  c.push_back(i);
  for (;;) {
    (i) = getchar_unlocked();
    if (i == ' ' || i == '\n' || i == '\r' || i == '\t' || i == EOF) break;
    c.push_back(i);
  };
  return ((int)c.size());
}
template <class T, class S>
void reader(T& x, S& y) {
  reader(x);
  reader(y);
}
template <class T, class S, class U>
void reader(T& x, S& y, U& z) {
  reader(x);
  reader(y);
  reader(z);
}
template <class T, class S, class U, class V>
void reader(T& x, S& y, U& z, V& w) {
  reader(x);
  reader(y);
  reader(z);
  reader(w);
}
inline void writer(int x, char c) {
  int s = 0, m = 0;
  char f[10];
  if (x < 0) m = 1, x = -x;
  while (x) f[s++] = x % 10, x /= 10;
  if (!s) f[s++] = 0;
  if (m) putchar_unlocked('-');
  while (s--) putchar_unlocked(f[s] + '0');
  putchar_unlocked(c);
}
inline void writer(ll x, char c) {
  int s = 0, m = 0;
  char f[20];
  if (x < 0) m = 1, x = -x;
  while (x) f[s++] = x % 10, x /= 10;
  if (!s) f[s++] = 0;
  if (m) putchar_unlocked('-');
  while (s--) putchar_unlocked(f[s] + '0');
  putchar_unlocked(c);
}
inline void writer(const char c[]) {
  int i;
  for (i = 0; c[i] != '\0'; i++) putchar_unlocked(c[i]);
}
inline void writer(const char x[], char c) {
  int i;
  for (i = 0; x[i] != '\0'; i++) putchar_unlocked(x[i]);
  putchar_unlocked(c);
}
template <class T>
void writerLn(T x) {
  writer(x, '\n');
}
template <class T, class S>
void writerLn(T x, S y) {
  writer(x, ' ');
  writer(y, '\n');
}
template <class T, class S, class U>
void writerLn(T x, S y, U z) {
  writer(x, ' ');
  writer(y, ' ');
  writer(z, '\n');
}
template <class T>
void writerArr(T x[], int n) {
  if (!n) {
    putchar_unlocked('\n');
    return;
  }
  for (int i = 0; i < (n - 1); i++) writer(x[i], ' ');
  writer(x[n - 1], '\n');
}
template <class T>
void writerArr(vector<T>& x) {
  writerArr(x.data(), (int)x.size());
}
const int MOD = ((int)1e9) + 7;
const int N = ((int)1e6) + 1;
int factors[N];
int dp[((int)1e6) + 1][21];
inline void init() {
  for (int i = 2; i < N; i += 2) factors[i] = 2;
  for (int i = 3; i < N; i += 2) factors[i] = i;
  for (int i = 3; i < N; i += 2)
    if (factors[i] == i)
      for (int j = 3 * i; j < N; j += 2 * i) factors[j] = i;
  dp[0][0] = 1;
  for (int i = 0; i < (20); i++) dp[0][i + 1] = 2;
  for (int j = 1; j <= ((int)1e6); j++) {
    dp[j][0] = 1;
    for (int k = 0; k <= 20; k++) {
      int tmp = dp[j][k] + dp[j - 1][k + 1];
      if (tmp >= MOD) tmp -= MOD;
      dp[j][k + 1] = tmp;
    }
  }
}
int main() {
  init();
  int q;
  reader(q);
  for (int i = 0; i < (q); i++) {
    int r, n;
    reader(r, n);
    ll ans = 1;
    while (n != 1) {
      int p = factors[n];
      int cnt = 0;
      while (n % p == 0) n /= p, cnt++;
      ans = ans * dp[r][cnt] % MOD;
    }
    writerLn(ans);
  }
  return 0;
}