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.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static int MOD = 1000000007;

    // After writing solution, quick scan for:
    //   array out of bounds
    //   special cases e.g. n=1?
    //   npe, particularly in maps
    //
    // Big numbers arithmetic bugs:
    //   int overflow
    //   sorting, or taking max, after MOD
    void solve() throws IOException {
        int[] nT = ril(2);
        int n = nT[0];
        int T = nT[1];
        int[][] tg = new int[n][];
        for (int i = 0; i < n; i++) tg[i] = ril(2);

        // dp[t][m][g] is the number of sequences possible to be formed from
        // songs indicated by mask m, having total time t, and whose last song
        // is genre g.
        int[][][] dp = new int[T+1][1 << n][3];
        for (int ti = 1; ti <= T; ti++) {
            for (int mask = 1; mask <= (1 << n); mask++) {
                // Iterate over the possible last song. It's genre will determine
                // what the third dimension should be
                for (int j = 0; j < n; j++) {
                    if (((1 << j) & mask) == 0) continue;
                    int time = tg[j][0];
                    int genre = tg[j][1]-1;
                    if (ti - time < 0) continue;
                    long base = dp[ti][mask][genre];
                    long add = 0;
                    if (ti - time == 0) {
                        // Base case: there is always one way regardless of what songs are available.
                        // The empty playlist
                        add = 1;
                    } else {
                        for (int g = 0; g < 3; g++) add += dp[ti-time][mask ^ (1 << j)][g];
                        add -= dp[ti-time][mask ^ (1 << j)][genre];
                    }
                    dp[ti][mask][genre] = (int) ((base + add) % MOD);
                }
            }
        }
        long ans = 0;
        for (int g = 0; g < 3; g++) ans += dp[T][(1 << n)-1][g];
        pw.println(ans % MOD);
    }
    // IMPORTANT
    // DID YOU CHECK THE COMMON MISTAKES ABOVE?

    // Template code below

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.solve();
        m.close();
    }

    void close() throws IOException {
        pw.flush();
        pw.close();
        br.close();
    }

    int ri() throws IOException {
        return Integer.parseInt(br.readLine().trim());
    }

    long rl() throws IOException {
        return Long.parseLong(br.readLine().trim());
    }

    int[] ril(int n) throws IOException {
        int[] nums = new int[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            int x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    long[] rll(int n) throws IOException {
        long[] nums = new long[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            long x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    int[] rkil() throws IOException {
        int sign = 1;
        int c = br.read();
        int x = 0;
        if (c == '-') {
            sign = -1;
            c = br.read();
        }
        while (c >= '0' && c <= '9') {
            x = x * 10 + c - '0';
            c = br.read();
        }
        return ril(x);
    }

    long[] rkll() throws IOException {
        int sign = 1;
        int c = br.read();
        int x = 0;
        if (c == '-') {
            sign = -1;
            c = br.read();
        }
        while (c >= '0' && c <= '9') {
            x = x * 10 + c - '0';
            c = br.read();
        }
        return rll(x);
    }

    char[] rs() throws IOException {
        return br.readLine().toCharArray();
    }

    void sort(int[] A) {
        Random r = new Random();
        for (int i = A.length-1; i > 0; i--) {
            int j = r.nextInt(i+1);
            int temp = A[i];
            A[i] = A[j];
            A[j] = temp;
        }
        Arrays.sort(A);
    }

    void printDouble(double d) {
        pw.printf("%.16f", d);
    }
}