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;
struct Tree {
  int l;
  int r;
  Tree *L{};
  Tree *R{};
  int val{};
  long long add{};
  long long sum{};
  Tree(int l, int r) : l(l), r(r) {
    if (l == r - 1) {
      val = l;
    } else {
      const auto m = (l + r) >> 1;
      L = new Tree(l, m);
      R = new Tree(m, r);
    }
  }
  void maintain() {
    val = L->val == R->val ? L->val : 0;
    sum = L->sum + R->sum;
  }
  void update(int _val, long long _add) {
    if (_val) {
      val = _val;
    }
    if (_add) {
      add += _add;
      sum += _add * (r - l);
    }
  }
  void pushdown() {
    L->update(val, add);
    R->update(val, add);
    add = 0;
  }
  void modify(int t, int v, int x) {
    if (v <= l || r <= t) return;
    if (t <= l && r <= v && val) {
      update(x, abs(val - x));
      return;
    }
    pushdown();
    L->modify(t, v, x);
    R->modify(t, v, x);
    maintain();
  }
  long long query(int t, int v) {
    if (v <= l || r <= t) return 0;
    if (t <= l && r <= v) return sum;
    pushdown();
    const auto x = L->query(t, v);
    const auto y = R->query(t, v);
    return x + y;
  }
};
int main() {
  int n, m;
  cin >> n >> m;
  Tree root(1, n + 1);
  while (m--) {
    int c;
    scanf("%d", &c);
    switch (c) {
      case 1: {
        int l, r, x;
        scanf("%d%d%d", &l, &r, &x);
        root.modify(l, r + 1, x);
        break;
      }
      case 2: {
        int l, r;
        scanf("%d%d", &l, &r);
        const auto ans = root.query(l, r + 1);
        printf("%lld\n", ans);
        break;
      }
    }
  }
  return 0;
}
