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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class BashPlaysFunctions {
    
    int N = (int) (1e6 + 5), M = 25;
    int SQRT = (int) (Math.sqrt(N) + 0.5);
    int MOD = (int) (1e9 + 7);
    
    void solve() {
        int[] fp = new int[N];
        for (int i = 2; i < N; i++) {
            if (fp[i] == 0) {
                fp[i] = i;
                for (int j = i * i; j < N && i <= SQRT; j += i) fp[j] = i;
            }
        }
        
        int[][] dp = new int[N][M];
        for (int i = 0; i < N; i++) dp[i][0] = 1;
        for (int j = 0; j < M; j++) dp[0][j] = 2;
        for (int i = 1; i < N; i++) {
            for (int j = 1; j < M; j++) {
                dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % MOD;
            }
        }
        
        int T = in.nextInt();
        while (T-- > 0) {
            int r = in.nextInt(), n = in.nextInt();
            long ans = 1;
            while (n > 1) {
                int x = fp[n], cnt = 0;
                while (n % x == 0) {
                    n /= x;
                    cnt++;
                }
                ans = ans * dp[r][cnt] % MOD;
            }
            out.println(ans);
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new BashPlaysFunctions().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}