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.
#include <bits/stdc++.h>
using namespace std;
long long N, M, ql, qr;
long long newC;
struct Node {
  long long col;
  long long value;
  long long lazy;
  long long lazyCol;
  Node(long long col = 0) : col(col), lazy(0), value(0), lazyCol(0) {}
  Node(Node a, Node b) {
    lazy = col = lazyCol = 0;
    if (a.col == b.col and a.col != 0) col = a.col;
    value = a.value + b.value;
  }
} T[111111 * 4];
void build(long long x, long long l, long long r) {
  if (l == r) {
    T[x] = Node(l);
    return;
  }
  long long m = (l + r) / 2;
  build(2 * x, l, m);
  build(2 * x + 1, m + 1, r);
  T[x] = Node(T[2 * x], T[2 * x + 1]);
}
void split(int x, int l, int r) {
  if (T[x].lazy) {
    T[x].value += T[x].lazy * (r - l + 1LL);
    if (l != r) {
      T[2 * x].lazy += T[x].lazy;
      T[2 * x + 1].lazy += T[x].lazy;
    }
    T[x].lazy = 0;
  }
  if (T[x].lazyCol) {
    T[x].col = T[x].lazyCol;
    if (l != r) {
      T[2 * x].lazyCol = T[x].lazyCol;
      T[2 * x + 1].lazyCol = T[x].lazyCol;
    }
    T[x].lazyCol = 0;
  }
}
void update(long long x, long long l, long long r) {
  split(x, l, r);
  if (qr < l or r < ql or qr < ql) return;
  if (ql <= l and r <= qr) {
    if (T[x].col == 0) {
      long long m = (l + r) / 2;
      update(2 * x, l, m);
      update(2 * x + 1, m + 1, r);
      T[x] = Node(T[2 * x], T[2 * x + 1]);
    } else {
      T[x].value += 1LL * abs(newC - T[x].col) * (r - l + 1LL);
      if (l != r) {
        T[2 * x].lazy += abs(newC - T[x].col);
        T[2 * x + 1].lazy += abs(newC - T[x].col);
        T[2 * x + 1].lazyCol = T[2 * x].lazyCol = newC;
      }
      T[x].col = newC;
    }
    return;
  }
  long long m = (l + r) / 2;
  update(2 * x, l, m);
  update(2 * x + 1, m + 1, r);
  T[x] = Node(T[2 * x], T[2 * x + 1]);
}
long long query(long long x, long long l, long long r) {
  if (qr < l or r < ql or qr < ql) return 0;
  split(x, l, r);
  if (ql <= l and r <= qr) return T[x].value;
  long long m = (l + r) / 2;
  return query(2 * x, l, m) + query(2 * x + 1, m + 1, r);
}
int main() {
  cin >> N >> M;
  build(1, 1, N);
  long long t;
  while (M--) {
    cin >> t >> ql >> qr;
    if (t == 1) {
      cin >> newC;
      update(1, 1, N);
    } else
      cout << query(1, 1, N) << endl;
  }
}
