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 JAVA):

Solution:
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;

public class D2 {
    FastScanner in;
    PrintWriter out;
    public static final int MAX = 1_000_010;
    public static final int[] sieve = new int[MAX];
    public final static int[][] precalc = new int[MAX][21];
    public static final int MOD = 1_000_000_007;

    static {
        for (int i = 2; i < MAX; i++) {
            if (sieve[i] == 0) {
                for (int j = i; j < MAX; j+=i) {
                    sieve[j] = i;
                }
            }
        }
        Arrays.fill(precalc[0], 2);
        precalc[0][0] = 1;
        for (int i = 1; i < precalc.length; i++) {
            precalc[i][0] = 1;
            for (int j = 1; j < precalc[0].length; j++) {
                precalc[i][j] = precalc[i][j - 1] + precalc[i - 1][j];
                if (precalc[i][j] >= MOD) {
                    precalc[i][j] -= MOD;
                }
            }
        }
    }

    List<Integer> get(int n) {
        List<Integer> ans = new ArrayList<>();
        while (n > 1) {
            int x = sieve[n];
            int cnt = 0;
            while (n % x == 0) {
                n /= x;
                cnt++;
            }
            ans.add(cnt);
        }
        return ans;
    }

    public void solve() throws IOException {
        int q = in.nextInt();
        for (int i = 0; i < q; i++) {
            int r = in.nextInt();
            List<Integer> n = get(in.nextInt());
            long ans = 1;
            for (int x : n) {
                ans *= precalc[r][x];
                ans %= MOD;
            }
            out.println(ans);
        }
    }

    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);
            Locale.setDefault(Locale.US);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
    }

    public static void main(String[] arg) {
        new D2().run();
    }
}