Problem: There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N


Output

Print the maximum total reward that you can earn no later than M days from today.

Examples

Input

3 4
4 3
4 1
2 2


Output

5


Input

5 3
1 2
1 3
1 4
2 1
2 3


Output

10


Input

1 1
2 1


Output

0

Solution (in JAVA):

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

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 */
public class Main {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static int MOD = 1000000007;

    private static void solve() throws IOException {
        int n = nextInt();
        int m = nextInt();

        int a[] = new int[n];
        int b[] = new int[n];
        Integer order[] = new Integer[n];
        for (int i = 0; i < n; i++) {
            order[i] = i;
            a[i] = nextInt();
            b[i] = nextInt();
        }
        Arrays.sort(order, (o1, o2) -> {
            if (b[o1] == b[o2]) {
                return a[o1] - a[o2];
            }
            return b[o2] - b[o1];
        });

        boolean vis[] = new boolean[m];
        long res = 0;
        int pre[] = new int[m + 1];
        pre[0] = -1;
        for (int i = 1; i < m; i++) {
            pre[i] = i - 1;
        }

        int s = 0;
        for (int j = 0; j < n && s < m; j++) {
            int i = order[j];
            int cur = m - a[i];
            while (cur >= 0 && vis[cur]) {
                cur = pre[cur];
            }
            if (cur < 0) {
                continue;
            }
            vis[cur] = true;
            res += b[i];
            pre[m - a[i]] = cur - 1;
            pre[cur + 1] = cur - 1;
        }
        pw.println(res);
    }

    static void swap(long a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void swap(int a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        solve();
        pw.flush();
    }

    static String next() throws IOException {
        st.nextToken();
        return st.sval;
    }

    static String next(int len) throws IOException {
        int b = br.read();
        while (b == '\n' || b == ' ' || b == '\r' || b == '\t') {
            b = br.read();
        }
        char ch[] = new char[len];
        ch[0] = (char) b;
        int idx = 1;
        while (idx < len && (b = br.read()) != ' ' && b != '\n' && b != '\r' && b != '\t') {
            ch[idx++] = (char) b;
        }
        return String.valueOf(ch).trim();
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

    private static String nextLine() throws IOException {
        return br.readLine();
    }
}