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 bufSize = 1 << 12;
const int mod = (int)1e9 + 7;
char buf[bufSize] = {};
int bufPos = bufSize;
inline char nextChar() {
  if (bufPos == bufSize) {
    memset(buf, '\0', bufSize);
    fread(buf, sizeof(char), bufSize, stdin);
    bufPos = 0;
  }
  return buf[bufPos++];
}
inline int nextInt() {
  char c = nextChar();
  while (c <= ' ') c = nextChar();
  int sign = +1, num = 0;
  if (c == '-' || c == '+') {
    sign = (c == '-') ? -1 : +1;
    c = nextChar();
  }
  while (c > ' ') {
    num = (num * 10) + c - '0';
    c = nextChar();
  }
  return num * sign;
}
const int maxNum = 1000000;
int divs[maxNum + 1];
int dp[33], pdp[33];
int curLayer = 0;
int n = nextInt();
vector<pair<pair<int, int>, int> > queries(n);
vector<int> ans(n);
inline void nextLayer() {
  for (int i = 0; i <= 32; i++) {
    pdp[i] = dp[i];
  }
  dp[0] = 1;
  for (int i = 1; i <= 32; i++) {
    dp[i] = pdp[i] + dp[i - 1];
    if (dp[i] >= mod) {
      dp[i] -= mod;
    }
  }
  curLayer++;
}
signed main() {
  dp[0] = 1;
  for (int i = 1; i <= 32; i++) {
    dp[i] = 2;
  }
  for (int i = 2; i <= maxNum; i++) {
    for (int j = i; j <= maxNum; j += i) {
      if (divs[j] == 0) {
        divs[j] = i;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    queries[i].first.first = nextInt();
    queries[i].first.second = nextInt();
    queries[i].second = i;
  }
  sort(queries.begin(), queries.end());
  for (int i = 0; i < n; i++) {
    while (queries[i].first.first != curLayer) {
      nextLayer();
    }
    int res = 1;
    int cur = queries[i].first.second;
    while (cur != 1) {
      int curCnt = 0;
      int curDiv = divs[cur];
      while (cur % curDiv == 0) {
        cur /= curDiv;
        curCnt++;
      }
      res = ((long long)res * dp[curCnt]) % mod;
    }
    ans[queries[i].second] = res;
  }
  for (int i = 0; i < n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}