Problem: Vasya has recently bought some land and decided to surround it with a wooden fence.

He went to a company called "Wooden board" that produces wooden boards for fences. Vasya read in the catalog of products that the company has at its disposal n different types of wood. The company uses the i-th type of wood to produce a board of this type that is a rectangular ai by bi block.

Vasya decided to order boards in this company and build a fence from them. It turned out that the storehouse of the company is so large that Vasya can order arbitrary number of boards of every type. Note that Vasya is allowed to turn the boards as he builds the fence. However, Vasya cannot turn square boards.

Vasya is required to construct a fence of length l, however, an arbitrary fence won't do. Vasya wants his fence to look beautiful. We'll say that a fence is beautiful if and only if the following two conditions are fulfilled:

  * there are no two successive boards of the same type 
  * the first board of the fence has an arbitrary length, and the length of each subsequent board equals the width of the previous one 



In other words, the fence is considered beautiful, if the type of the i-th board in the fence is different from the i - 1-th board's type; besides, the i-th board's length is equal to the i - 1-th board's width (for all i, starting from 2).

Now Vasya wonders, how many variants of arranging a fence for his land exist. Your task is to count the number of different beautiful fences of length l.

Two fences will be considered the same if the corresponding sequences of fence boards types and rotations are the same, otherwise the fences are different. Since the sought number can be large enough, you need to calculate the answer modulo 1000000007 (109 + 7).

Input

The first line contains two integers n and l (1 ≤ n ≤ 100, 1 ≤ l ≤ 3000) — the number of different board types and the fence length, correspondingly. Next n lines contain descriptions of board types: the i-th line contains two integers ai and bi (1 ≤ ai, bi ≤ 100) — the sizes of the board of the i-th type. All numbers on the lines are separated by spaces.

Output

Print a single integer — the sought number of variants modulo 1000000007 (109 + 7).

Examples

Input

2 3
1 2
2 3


Output

2


Input

1 2
2 2


Output

1


Input

6 6
2 1
3 2
2 5
3 3
5 1
2 1


Output

20

Note

In the first sample there are exactly two variants of arranging a beautiful fence of length 3: 

  * As the first fence board use the board of the first type of length 1 and width 2. As the second board use board of the second type of length 2 and width 3. 
  * Use one board of the second type after you turn it. That makes its length equal 3, and width — 2. 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

// 4/24/12 8:44 PM
// by Abrackadabra

public class E {
    public static void main(String[] args) throws IOException {
        if (args.length > 0 && args[0].equals("Abra")) debugMode = true;
        new E().run();
        /*new Thread(null, new Runnable() {
            public void run() {
                try {
                    new Test().run();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }, "1", 1 << 24).start();*/
    }

    int IOMode = 0; //0 - consoleIO, 1 - <taskName>.in/out, 2 - input.txt/output.txt, 3 - test case generator
    String taskName = "";

    void solve() throws IOException {
        int MOD = 1000000007;
        int n = nextInt(), l = nextInt();
        int[] a = new int[n], b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            b[i] = nextInt();
        }
        int[][][] dp = new int[l + 200][n][2];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < (a[i] == b[i] ? 1 : 2); j++) {
                int l0 = j == 0 ? a[i] : b[i];
                dp[l0][i][j] = 1;
            }
        }
        for (int i = 1; i < l; i++) {
            for (int j0 = 0; j0 < n; j0++) {
                for (int r0 = 0; r0 < (a[j0] == b[j0] ? 1 : 2); r0++) {
                    for (int j1 = 0; j1 < n; j1++) {
                        for (int r1 = 0; r1 < (a[j1] == b[j1] ? 1 : 2); r1++) {
                            if (j0 == j1) continue;
                            int l0 = a[j0];
                            int w0 = b[j0];
                            if (r0 == 1) {l0 = b[j0]; w0 = a[j0];}
                            int l1 = a[j1];
                            int w1 = b[j1];
                            if (r1 == 1) {l1 = b[j1]; w1 = a[j1];}
                            if (w0 == l1 && i + l1 <= l) {
                                dp[i + l1][j1][r1] = (dp[i + l1][j1][r1] + dp[i][j0][r0]) % MOD;
                            }
                        }
                    }
                }
            }
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 2; j++) {
                res = (res + dp[l][i][j]) % MOD;
            }
        }
        out.println(res);
    }

    long startTime = System.nanoTime(), tempTime = startTime, finishTime = startTime;
    long startMem = Runtime.getRuntime().totalMemory(), finishMem = startMem;

    void run() throws IOException {
        init();
        if (debugMode) {
            con.println("Start");
            con.println("Console:");
        }
        solve();
        finishTime = System.nanoTime();
        finishMem = Runtime.getRuntime().totalMemory();
        out.flush();
        if (debugMode) {
            int maxSymbols = 1000;
            BufferedReader tbr = new BufferedReader(new FileReader("input.txt"));
            char[] a = new char[maxSymbols];
            tbr.read(a);
            if (a[0] != 0) {
                con.println("File input:");
                con.println(a);
                if (a[maxSymbols - 1] != 0) con.println("...");
            }
            tbr = new BufferedReader(new FileReader("output.txt"));
            a = new char[maxSymbols];
            tbr.read(a);
            if (a[0] != 0) {
                con.println("File output:");
                con.println(a);
                if (a[maxSymbols - 1] != 0) con.println("...");
            }
            con.println("Execution time: " + (finishTime - startTime) / 1000000000.0 + " sec");
            con.println("Used memory:    " + (finishMem - startMem) + " bytes");
            con.println("Total memory:   " + Runtime.getRuntime().totalMemory() + " bytes");
        }
    }

    boolean tick(double x) {
        if (System.nanoTime() - tempTime > x * 1e9) {
            tempTime = System.nanoTime();
            con.println("Tick at " + (tempTime - startTime) / 1000000000 + " sec");
            con.print("   ");
            return true;
        }
        return false;
    }

    static boolean debugMode = false;
    PrintStream con = System.out;

    void init() throws IOException {
        if (debugMode && IOMode != 3) {
            br = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter(new FileWriter("output.txt"));
        } else
            switch (IOMode) {
                case 0:
                    br = new BufferedReader(new InputStreamReader(System.in));
                    out = new PrintWriter(System.out);
                    break;
                case 1:
                    br = new BufferedReader(new FileReader(taskName + ".in"));
                    out = new PrintWriter(new FileWriter(taskName + ".out"));
                    break;
                case 2:
                    br = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter(new FileWriter("output.txt"));
                    break;
                case 3:
                    out = new PrintWriter(new FileWriter("input.txt"));
                    break;
            }
        if (!debugMode)
            con = new PrintStream(new OutputStream() {
                public void write(int b) throws IOException {
                } // to /dev/null
            });
        if ((IOMode == 1 && taskName.length() == 0) || (IOMode != 1 && taskName.length() != 0)) {
            System.err.println("Leha, check IO settings");
            System.err.flush();
            System.exit(0);
        }
    }

    BufferedReader br;
    PrintWriter out;
    StringTokenizer in;

    boolean hasMoreTokens() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            String line = br.readLine();
            if (line == null) return false;
            in = new StringTokenizer(line);
        }
        return true;
    }

    String nextString() throws IOException {
        return hasMoreTokens() ? in.nextToken() : null;
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextString());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextString());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }
}