Problem: Anton likes permutations, especially he likes to permute their elements. Note that a permutation of n elements is a sequence of numbers {a1, a2, ..., an}, in which every number from 1 to n appears exactly once.

One day Anton got a new permutation and started to play with it. He does the following operation q times: he takes two elements of the permutation and swaps these elements. After each operation he asks his friend Vanya, how many inversions there are in the new permutation. The number of inversions in a permutation is the number of distinct pairs (i, j) such that 1 ≤ i < j ≤ n and ai > aj.

Vanya is tired of answering Anton's silly questions. So he asked you to write a program that would answer these questions instead of him.

Initially Anton's permutation was {1, 2, ..., n}, that is ai = i for all i such that 1 ≤ i ≤ n.

Input

The first line of the input contains two integers n and q (1 ≤ n ≤ 200 000, 1 ≤ q ≤ 50 000) — the length of the permutation and the number of operations that Anton does.

Each of the following q lines of the input contains two integers li and ri (1 ≤ li, ri ≤ n) — the indices of elements that Anton swaps during the i-th operation. Note that indices of elements that Anton swaps during the i-th operation can coincide. Elements in the permutation are numbered starting with one.

Output

Output q lines. The i-th line of the output is the number of inversions in the Anton's permutation after the i-th operation.

Examples

Input

5 4
4 5
2 4
2 5
2 2


Output

1
4
3
3


Input

2 1
2 1


Output

1


Input

6 7
1 4
3 5
2 3
3 3
3 6
2 1
5 1


Output

5
6
7
7
10
11
8

Note

Consider the first sample.

After the first Anton's operation the permutation will be {1, 2, 3, 5, 4}. There is only one inversion in it: (4, 5).

After the second Anton's operation the permutation will be {1, 5, 3, 2, 4}. There are four inversions: (2, 3), (2, 4), (2, 5) and (3, 4).

After the third Anton's operation the permutation will be {1, 4, 3, 2, 5}. There are three inversions: (2, 3), (2, 4) and (3, 4).

After the fourth Anton's operation the permutation doesn't change, so there are still three inversions.

Solution (in JAVA):

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

public class E {
	public static int sqrt, maxn, bit[][], arr[];

	public static void main(String[] args) throws Exception {
		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
		
		int n = in.nextInt();
		int q = in.nextInt();

		sqrt = 1 + (int) Math.sqrt(n);
		maxn = 200000 + 5;
		bit = new int[sqrt + 5][maxn];

		arr = new int[maxn];

		long ans = 0;

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

		for (int i = 1; i <= n; i++) {
			int j = i / sqrt;

			if (i % sqrt != 0) {
				j++;
			}

			update(j, arr[i], 1);
		}

		for (int qq = 1; qq <= q; qq++) {
			int l1 = in.nextInt(), r1 = in.nextInt();

			if (l1 > r1) {
				int t = l1;
				l1 = r1;
				r1 = t;
			}

			int ls1 = l1 / sqrt, rs1 = r1 / sqrt;

			if (l1 % sqrt != 0) {
				ls1++;
			}

			if (r1 % sqrt != 0) {
				rs1++;
			}

			int l = l1 + 1;
			int r = r1 - 1;

			int ls = l / sqrt;
			int rs = r / sqrt;

			if (l % sqrt != 0) {
				ls++;
			}

			if (r % sqrt != 0) {
				rs++;
			}

			ans -= getBigger(l, r, ls, rs, n, arr[r1]);
			ans -= getSmaller(l, r, ls, rs, n, arr[l1]);
			ans += getBigger(l, r, ls, rs, n, arr[l1]);
			ans += getSmaller(l, r, ls, rs, n, arr[r1]);

			if (arr[l1] > arr[r1]) {
				ans--;
			} else if (arr[r1] > arr[l1]) {
				ans++;
			}
			
			update(ls1, arr[l1], -1);
			update(rs1, arr[l1], 1);
			update(ls1, arr[r1], 1);
			update(rs1, arr[r1], -1);

			int temp = arr[l1];
			arr[l1] = arr[r1];
			arr[r1] = temp;

			out.println(ans);
		}

		out.close();
	}

	static int getBigger(int l, int r, int ls, int rs, int n, int num) {
		int cnt = 0;
		
		if (rs - ls <= 1) {
			for (int i = l; i <= r; i++) {
				if (arr[i] > num) {
					cnt++;
				}
			}
			
			return cnt;
		} else {
			cnt += query(rs - 1, num + 1, n);
			cnt -= query(ls, num + 1, n);
			
			for (int i = l; i <= (ls) * sqrt; i++) {
				if (arr[i] > num) {
					cnt++;
				}
			}

			for (int i = (rs - 1) * sqrt + 1; i <= r; i++) {
				if (arr[i] > num) {
					cnt++;
				}
			}
			
			return cnt;
		}
	}

	static int getSmaller(int l, int r, int ls, int rs, int n, int num) {
		int cnt = 0;
		
		if (rs - ls <= 1) {
			for (int i = l; i <= r; i++) {
				if (arr[i] < num) {
					cnt++;
				}
			}
			
			return cnt;
		} else {
			cnt += query(rs - 1, 1, num - 1);
			cnt -= query(ls, 1, num - 1); 
			
			for (int i = l; i <= (ls) * sqrt; i++) {
				if (arr[i] < num) {
					cnt++;
				}
			}

			for (int i = (rs - 1) * sqrt + 1; i <= r; i++) {
				if (arr[i] < num) {
					cnt++;
				}
			}
			
			return cnt;
		}

	}

	public static void update(int x, int y, int val) {
		int y1;
		
		while (x < bit.length) {
			y1 = y;
		
			while (y1 < bit[x].length) {
				bit[x][y1] += val;
				y1 += (y1 & -y1);
			}
			
			x += (x & -x);
		}
	}

	public static int getSum(int x, int y) {
		int sum = 0;
		
		while (x > 0) {
			int y1 = y;
		
			while (y1 > 0) {
				sum += bit[x][y1];
				y1 -= y1 & -y1;
			}
			
			x -= x & -x;
		}
		
		return sum;
	}

	public static int query(int x, int l, int r) {
		int p = getSum(x, r);
		int q = getSum(x, l - 1);

		return p - q;
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() throws Exception {
			while (tokenizer == null || !tokenizer.hasMoreTokens())
				tokenizer = new StringTokenizer(reader.readLine());
			return tokenizer.nextToken();
		}

		public String nextLine() throws Exception {
			String line = null;
			tokenizer = null;
			line = reader.readLine();
			return line;
		}

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

		public double nextDouble() throws Exception {
			return Double.parseDouble(next());
		}

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