DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

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


Output

8


Input

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


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.Scanner;

public class A {

	final int MAX_ID = 1 << 17;
	final int NON_UNIFORM = -1;
	int[] t_Color = new int[MAX_ID * 2];
	long[] t_Sum = new long[MAX_ID * 2];
	long[] t_Uni = new long[MAX_ID * 2];

	public static void main(String[] args) {
		new A().solve();
	}

	void solve() {
		Scanner scan = new Scanner(new BufferedInputStream(System.in));
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

		scan.nextInt(); // n
		int m = scan.nextInt();

		treeInit();

		for (int i = 0; i < m; i++) {
			int type = scan.nextInt();
			int l = scan.nextInt();
			int r = scan.nextInt();
			if (type == 1) {
				int x = scan.nextInt();
				update(l, r + 1, x);
			} else {
				out.println(quary(l, r + 1));
			}
		}

		out.flush();

		scan.close();
		out.close();
	}

	void treeInit() {
		treeInit(0, 0, MAX_ID);
	}

	void treeInit(int id, int a, int b) {
		t_Sum[id] = t_Uni[id] = 0;
		if (a + 1 == b) {
			t_Color[id] = a;
		} else {
			int m = (a + b) / 2;
			treeInit(id * 2 + 1, a, m);
			treeInit(id * 2 + 2, m, b);
			t_Color[id] = NON_UNIFORM;
		}
	}

	void update(int l, int r, int x) {
		update(l, r, x, 0, 0, MAX_ID);
	}

	long quary(int l, int r) {
		return quary(l, r, 0, 0, MAX_ID);
	}

	long update(int l, int r, int x, int id, int a, int b) {

		long res = 0;

		if (b <= l || r <= a) {

		} else if (l <= a && b <= r) {
			if (t_Color[id] == NON_UNIFORM) {
				res = update(l, r, x, id * 2 + 1, a, (a + b) / 2);
				res += update(l, r, x, id * 2 + 2, (a + b) / 2, b);
				// System.out.printf("Sum add %d-%d %d\n", a, b, res);
				t_Sum[id] += res;
			} else {
				long add = Math.abs(x - t_Color[id]);
				t_Uni[id] += add;
				res = add * (b - a);
				// System.out.printf("Uni add %d-%d %d\n", a, b, res);
			}
			t_Color[id] = x;
		} else {
			if (t_Color[id] != NON_UNIFORM) {
				t_Color[id * 2 + 1] = t_Color[id * 2 + 2] = t_Color[id];
				t_Color[id] = NON_UNIFORM;
			}

			res = update(l, r, x, id * 2 + 1, a, (a + b) / 2);
			res += update(l, r, x, id * 2 + 2, (a + b) / 2, b);
			// System.out.printf("Sum add %d-%d %d\n", a, b, res);
			t_Sum[id] += res;
		}

		return res;
	}

	long quary(int l, int r, int id, int a, int b) {
		if (b <= l || r <= a) {
			return 0;
		} else if (l <= a && b <= r) {
			// System.out.printf("%d-%d - %d * %d + %d\n", a, b, t_Uni[id], b -
			// a, t_Sum[id]);
			return t_Uni[id] * (b - a) + t_Sum[id];
		} else {
			long res = (Math.min(b, r) - Math.max(a, l)) * t_Uni[id];
			// System.out.printf("%d-%d - %d\n", a, b, res);
			int m = (a + b) / 2;
			return res + quary(l, r, id * 2 + 1, a, m)
					+ quary(l, r, id * 2 + 2, m, b);
		}
	}
}
