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;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct node {
  int l, r;
  long long s, col;
  long long add;
  long long length() { return r - l + 1; }
} segtree[400000 * 5];
void up(int root) {
  segtree[root].s = segtree[root * 2 + 1].s + segtree[root * 2 + 2].s;
  if (segtree[root * 2 + 1].col != segtree[root * 2 + 2].col) {
    segtree[root].col = 0;
  } else
    segtree[root].col = segtree[root * 2 + 1].col;
}
void down(int root) {
  if (segtree[root].col) {
    segtree[root * 2 + 1].s +=
        segtree[root * 2 + 1].length() * segtree[root].add;
    segtree[root * 2 + 2].s +=
        segtree[root * 2 + 2].length() * segtree[root].add;
    segtree[root * 2 + 1].col = segtree[root].col;
    segtree[root * 2 + 2].col = segtree[root].col;
    segtree[root * 2 + 1].add += segtree[root].add;
    segtree[root * 2 + 2].add += segtree[root].add;
    segtree[root].add = 0;
    segtree[root].col = 0;
  }
}
void build(int root, int l, int r) {
  segtree[root].l = l, segtree[root].r = r;
  segtree[root].col = 0;
  if (l == r) {
    segtree[root].col = l;
    return;
  }
  int mid = (l + r) >> 1;
  build(root * 2 + 1, l, mid);
  build(root * 2 + 2, mid + 1, r);
  up(root);
}
void update(int root, int l, int r, long long val) {
  if (segtree[root].l >= l && segtree[root].r <= r) {
    if (segtree[root].col) {
      segtree[root].s += segtree[root].length() * fabs(val - segtree[root].col);
      segtree[root].add += fabs(val - segtree[root].col);
      segtree[root].col = val;
      return;
    }
  }
  down(root);
  if (segtree[root * 2 + 1].r >= l) update(root * 2 + 1, l, r, val);
  if (segtree[root * 2 + 2].l <= r) update(root * 2 + 2, l, r, val);
  up(root);
}
long long query(int root, int l, int r) {
  long long ans = 0;
  if (segtree[root].l >= l && segtree[root].r <= r) {
    ans += segtree[root].s;
    return ans;
  }
  down(root);
  if (segtree[root * 2 + 1].r >= l) ans += query(root * 2 + 1, l, r);
  if (segtree[root * 2 + 2].l <= r) ans += query(root * 2 + 2, l, r);
  return ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  build(1, 1, n);
  while (m--) {
    int op, l, r, val;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &l, &r, &val);
      update(1, l, r, val);
    } else {
      scanf("%d%d", &l, &r);
      cout << query(1, l, r) << '\n';
    }
  }
  return 0;
}
