Problem: The only difference between easy and hard versions is constraints.

Polycarp loves to listen to music, so he never leaves the player, even on the way home from the university. Polycarp overcomes the distance from the university to the house in exactly T minutes.

In the player, Polycarp stores n songs, each of which is characterized by two parameters: t_i and g_i, where t_i is the length of the song in minutes (1 ≤ t_i ≤ 15), g_i is its genre (1 ≤ g_i ≤ 3).

Polycarp wants to create such a playlist so that he can listen to music all the time on the way from the university to his home, and at the time of his arrival home, the playlist is over. Polycarp never interrupts songs and always listens to them from beginning to end. Thus, if he started listening to the i-th song, he would spend exactly t_i minutes on its listening. Polycarp also does not like when two songs of the same genre play in a row (i.e. successively/adjacently) or when the songs in his playlist are repeated.

Help Polycarpus count the number of different sequences of songs (their order matters), the total duration is exactly T, such that there are no two consecutive songs of the same genre in them and all the songs in the playlist are different.

Input

The first line of the input contains two integers n and T (1 ≤ n ≤ 15, 1 ≤ T ≤ 225) — the number of songs in the player and the required total duration, respectively.

Next, the n lines contain descriptions of songs: the i-th line contains two integers t_i and g_i (1 ≤ t_i ≤ 15, 1 ≤ g_i ≤ 3) — the duration of the i-th song and its genre, respectively.

Output

Output one integer — the number of different sequences of songs, the total length of exactly T, such that there are no two consecutive songs of the same genre in them and all the songs in the playlist are different. Since the answer may be huge, output it modulo 10^9 + 7 (that is, the remainder when dividing the quantity by 10^9 + 7).

Examples

Input


3 3
1 1
1 2
1 3


Output


6


Input


3 3
1 1
1 1
1 3


Output


2


Input


4 10
5 3
2 1
3 2
5 1


Output


10

Note

In the first example, Polycarp can make any of the 6 possible playlist by rearranging the available songs: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2] and [3, 2, 1] (indices of the songs are given).

In the second example, the first and second songs cannot go in succession (since they have the same genre). Thus, Polycarp can create a playlist in one of 2 possible ways: [1, 3, 2] and [2, 3, 1] (indices of the songs are given).

In the third example, Polycarp can make the following playlists: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1], [1, 4], [4, 1], [2, 3, 4] and [4, 3, 2] (indices of the songs are given).

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    int n, t;
    void solve(int TC) throws Exception{
        n = ni();t = ni();
        int[][] song = new int[n][];
        for(int i = 0; i< n; i++)song[i] = new int[]{ni(), ni()-1};
        long[][] dp = new long[1<<n][3];
        for(int i = 0; i< dp.length; i++)
            for(int j = 0; j< dp[i].length; j++)
                dp[i][j] = -1;
        pn(start(dp, song, 0, -1));
    }
    long start(long[][] dp, int[][] song, int mask, int prev){
        long ti = 0;
        for(int i = 0; i< n; i++){
            if(((mask>>i)&1)==1)ti+=song[i][0];
        }
        if(ti==t)return 1;
        if(prev != -1 && dp[mask][prev] != -1)return dp[mask][prev];
        long ans = 0;
        for(int i = 0; i< n; i++){
            if(((mask>>i)&1)==0 && song[i][1] != prev && ti+song[i][0] <= t)
                ans = (ans+start(dp, song, mask|(1<<i), song[i][1]))%mod;
        }
        if(prev!= -1)dp[mask][prev] = ans;
        return ans;
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    long mod = (long)1e9+7, IINF = (long)1e18;
    final int INF = (int)1e9, MX = (int)2e5+5;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
    static boolean multipleTC = false, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }
}