Alyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special. 

Alyona is a capricious girl so after she gets the array, she inspects m of its subarrays. Subarray is a set of some subsequent elements of the array. The i-th subarray is described with two integers li and ri, and its elements are a[li], a[li + 1], ..., a[ri].

Alyona is going to find mex for each of the chosen subarrays. Among these m mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible. 

You are to find an array a of n elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.

The mex of a set S is a minimum possible non-negative integer that is not in S.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105).

The next m lines contain information about the subarrays chosen by Alyona. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), that describe the subarray a[li], a[li + 1], ..., a[ri].

Output

In the first line print single integer — the maximum possible minimum mex.

In the second line print n integers — the array a. All the elements in a should be between 0 and 109.

It is guaranteed that there is an optimal answer in which all the elements in a are between 0 and 109.

If there are multiple solutions, print any of them.

Examples

Input

5 3
1 3
2 5
4 5


Output

2
1 0 2 1 0


Input

4 2
1 4
2 4


Output

3
5 2 0 1

Note

The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.

import java.io.*;
import java.text.DecimalFormat;
import java.util.*;
public class CF_739A {
	public static void main(String[] args) {
		new CF_739A().run();
	}
	public void run(){
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int[] a = new int[n];
		int m = sc.nextInt();
		
		int min = Integer.MAX_VALUE;
		for(int i = 0;i<m;i++){
			int l = sc.nextInt()-1;
			int r = sc.nextInt()-1;
			min = Math.min(min, r-l+1);
		}
		out.println(min);
		int cur = 0;
		for(int i = 0;i<n-1;i++){
			out.print(cur%min + " ");
			cur++;
		}
		out.println(cur%min);
		out.close();
	}
}

class FastScanner {
	public BufferedReader br;
	public StringTokenizer st;

	public FastScanner(InputStream in) {
		br = new BufferedReader(new InputStreamReader(in));
		st = null;
	}

	public boolean hasMoreTokens() {
		while (st == null || !st.hasMoreTokens()) {
			String line = null;
			try {
				line = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (line == null)
				return false;
			st = new StringTokenizer(line);
		}
		return true;
	}

	public String next() {
		if (!hasMoreTokens())
			return null;
		return st.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	//读一个int类型的数组
	public int[] readIntArray(int n) {
		int[] array = new int[n];
		for (int i = 0; i < n; i++) {
			array[i] = nextInt();
		}
		return array;
	}

	//读一个long数组
	public long[] readLongArray(int n) {
		long[] array = new long[n];
		for (int i = 0; i < n; i++) {
			array[i] = nextLong();
		}
		return array;
	}

	//读一个字符串数组
	public String[] readStringArray(int n) {
		String[] a = new String[n];
		for (int i = 0; i < n; i++) {
			a[i] = next();
		}
		return a;
	}

	//打乱一个数组，便于排序
	public void shuffleIntArray(int[] a) {
		Random rand = new Random();
		for (int i = 0; i < a.length; i++) {
			int x = rand.nextInt(i + 1); // 0~i
			int t = a[i];
			a[i] = a[x];
			a[x] = t;
		}
	}

	// 只适合比较小的数，太大的数会越界
	public long factorial(long n) {
		if (n == 0)
			return 1L;
		long res = 1;
		for (long i = 2; i <= n; i++) {
			res *= i;
		}
		return res;
	}

	// 只适合比较小的数，太大的数会越界
	public long combination(long mom, long son) {
		return factorial(mom) / (factorial(son) * factorial(mom - son));
	}

	//求最小公倍数
	public long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	//求最大公约数
	public long lcm(long a, long b) {
		return a * b / (gcd(a, b));
	}

	//格式化double类型的数
	public double format(String pattern, double d) {
		DecimalFormat df = new DecimalFormat(pattern);
		return Double.parseDouble(df.format(d));
	}

	//判断一个数是否是2的指数
	public boolean isPowerOfTwo(long v) {
		return (v & (v - 1)) == 0;
	}
}
